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Calculus I Limits and Derivatives

Sunil Karna

Section 4.1 Integrations

\begin{align*} \int (x^n)\, dx \amp = \frac{x^{n+1}}{n+1}+C; \, n\neq -1\\ \int \frac{1}{x}\, dx \amp = \ln |x| +C \\ \int (e^x)\, dx \amp = e^x+C \\ \int (a^x)\, dx \amp = \frac{a^x}{\ln a}+C \\ \int (a^x)\, dx \amp = \frac{a^x}{\ln a}+C \\ \int \sin x\, dx \amp = -\cos x +C \\ \int \cos x\, dx \amp = \sin x +C \\ \int \sec^2 x\, dx \amp = \tan x +C \\ \int \csc^2 x\, dx \amp = -\cot x +C \\ \int \sec x \tan x\, dx \amp = \sec x +C \end{align*}
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Subsection 4.1.1 Methods of Integration

There are basic four methods of doing integration:
  • Integration by Inspection Method:
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  • Integration by Substitution Method:
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  • Integration using Partial Fractions:
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  • Integration by Parts:
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Subsubsection 4.1.1.1 Integration by Inspection Method:

Example 4.1.1.
Find the antiderivative (or integral) of the following functions by inspection method.
  1. \begin{equation*} \sin 3x \end{equation*}
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  2. \begin{equation*} \cos 2x \end{equation*}
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  3. \begin{equation*} (ax+b)^2 \end{equation*}
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  4. \begin{equation*} e^{2x} \end{equation*}
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  5. \begin{equation*} \sin 2x-3e^{3x} \end{equation*}
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Solution.
  1. \begin{equation*} \int \sin 3x\, dx = -\frac{\cos 3x}{3}+C \end{equation*}
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    As
    \begin{equation*} \frac{d}{dx} (\cos 3x)= -\sin 3x\cdot 3 =-3\sin 3x \end{equation*}
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  2. \begin{equation*} \int \cos 2x\, dx = \frac{\sin 2x}{2}+C \end{equation*}
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  3. \begin{equation*} \int (ax+b)^2 \, dx = \frac{(ax+b)^3}{3a} +C \end{equation*}
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  4. \begin{equation*} \int e^{2x}\, dx = \frac{e^{2x}}{2}+C \end{equation*}
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  5. \begin{equation*} \int (\sin 2x-3e^{3x}) \, dx = \int \sin 2x \, dx -3 \int e^{3x} \, dx \end{equation*}
    \begin{equation*} \, = -\frac{\cos 2x}{2}-3\frac{e^{3x}}{3}+C= -\frac{\cos 2x}{2}-e^{3x}+C \end{equation*}
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Example 4.1.2.
Find the antiderivative (or integral) of the following functions by inspection method.
  1. \begin{equation*} \int (4e^{3x}+1)\, dx \end{equation*}
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  2. \begin{equation*} \int x^2(1-\frac{1}{x^2})\, dx \end{equation*}
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  3. \begin{equation*} \int (2x^2+e^x)\, dx \end{equation*}
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  4. \begin{equation*} \int (ax^2+bx+c) \, dx \end{equation*}
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  5. \begin{equation*} \int \frac{x^3+5x^2-4}{x^2} \, dx \end{equation*}
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  6. \begin{equation*} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \, dx \end{equation*}
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Solution.
  1. \begin{equation*} \int (4e^{3x}+1)\, dx = \int 4e^{3x}\, dx+\int dx =4\frac{e^{3x}}{3}+x+c \end{equation*}
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  2. \begin{equation*} \int x^2(1-\frac{1}{x^2})\, dx = \int (x^2-1)\, dx = \frac{x^3}{3}-x+C \end{equation*}
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  3. \begin{equation*} \int (2x^2+e^x)\, dx = \fillinmath{XXXXX} \end{equation*}
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  4. \begin{equation*} \int (ax^2+bx+c) \, dx =\fillinmath{XXXXX} \end{equation*}
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  5. \begin{equation*} \int \frac{x^3+5x^2-4}{x^2} \, dx = \fillinmath{XXXXX} \end{equation*}
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  6. \begin{equation*} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \, dx = \int \left(x + \frac{1}{x}-2\right) \, dx \end{equation*}
    \begin{equation*} \, = \int x \, dx +\int \frac{1}{x}\, dx -2\int \, dx = \frac{x^2}{2}+\log |x| -2x+C \end{equation*}
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Example 4.1.3.
Find the integration of
\begin{equation*} \int \frac{x^3+3x+5}{\sqrt{x}} \, dx \end{equation*}
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Solution.
\begin{equation*} \int \frac{x^3+3x+5}{\sqrt{x}} \, dx = \int \left(x^{3-\frac{1}{2}}+3 x^{1-\frac{1}{2}}+4x^{-\frac{1}{2}}\right) \, dx \end{equation*}
\begin{equation*} \, = \int \left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4x^{-\frac{1}{2}}\right) \, dx = \frac{2}{7}x^{\frac{7}{2}}+2x^{\frac{3}{2}} +8x^{\frac{1}{2}}+C \end{equation*}
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Example 4.1.4.
Find the integration of
  1. \begin{equation*} \int \frac{x^3-x^2+x-1+5}{x-1} \, dx \end{equation*}
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  2. \begin{equation*} \int (1-x) \sqrt{x} \, dx \end{equation*}
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  3. \begin{equation*} \int (2x-3\cos x +e^x) \, dx \end{equation*}
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  4. \begin{equation*} \int \sqrt{x}(3x^2+2x+3) \, dx \end{equation*}
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  5. \begin{equation*} \int (2x^2-3\sin x+5\sqrt{x}) \, dx \end{equation*}
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Solution.
  1. \begin{equation*} \int \frac{x^3-x^2+x-1+5}{x-1} \, dx = \int \frac{(x^2+1)(x-1)}{(x-1)} \, dx \end{equation*}
    \begin{equation*} =\int (x^2+1) \, dx= \frac{x^3}{3} +x+C \end{equation*}
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  2. \begin{equation*} \int (1-x) \sqrt{x} \, dx = \int \sqrt{x} \, dx -\int x\sqrt{x} \, dx = \frac{2}{3}x^{\frac{3}{2}}-\frac{2}{5}x^{\frac{5}{2}}+C \end{equation*}
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  3. \begin{equation*} \int (2x-3\cos x +e^x) \, dx = \int 2x \, dx -3\int \cos x \, dx +\int e^x \, dx \end{equation*}
    \begin{equation*} = x^2 -3\sin x +e^x+C \end{equation*}
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  4. \begin{equation*} \int \sqrt{x}(3x^2+2x+3) \, dx = \int (3x^{\frac{5}{2}}+2x^{\frac{3}{2}}+3x^{\frac{1}{2}}) \, dx \end{equation*}
    \begin{equation*} = \frac{6}{7}x^{\frac{7}{2}}+\frac{4}{5}x^{\frac{5}{2}}+2x^{\frac{3}{2}}+C \end{equation*}
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  5. \begin{equation*} \int (2x^2-3\sin x+5\sqrt{x}) \, dx = \int 2x^2 \, dx -3\int \sin x \, dx +5\int \sqrt{x} \, dx \end{equation*}
    \begin{equation*} = \frac{2}{3}x^3 +3\cos x +\frac{10}{3}x^{\frac{3}{2}}+C \end{equation*}
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Example 4.1.5.
Find the integration of
  1. \begin{equation*} \int \sec x (\sec x+\tan x)\, dx \end{equation*}
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  2. \begin{equation*} \int \frac{\sec^2 x}{\csc^2 x} \, dx \end{equation*}
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  3. \begin{equation*} \int \frac{2-3\sin x}{\cos^2 x} \, dx \end{equation*}
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  4. \begin{equation*} \int \sqrt{x}(3x^2+2x+3) \, dx \end{equation*}
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Solution.
  1. \begin{equation*} \int \sec x (\sec x+\tan x)\, dx = \int (\sec^2 x+\sec x\tan x)\, dx \end{equation*}
    \begin{equation*} = \tan x +\sec x+C \end{equation*}
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  2. \begin{equation*} \int \frac{\sec^2 x}{\csc^2 x} \, dx = \end{equation*}
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  3. \begin{equation*} \int \frac{2-3\sin x}{\cos^2 x} \, dx = \int 2\sec^2 x \, dx -3 \int \sec x\tan x\, dx \end{equation*}
    \begin{equation*} = 2\tan x -3\sec x+C \end{equation*}
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Example 4.1.6.
If \(\frac{d}{dx}f(x) = 4x^3- \frac{3}{x^4}\) such that \(f(2) = 0\text{,}\) then find \(f(x).\)
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Solution.
\begin{equation*} \frac{d}{dx}f(x) = 4x^3- \frac{3}{x^4} \end{equation*}
\begin{equation*} or, \, \int d[f(x)] = \int (4x^3- \frac{3}{x^4})\, dx \end{equation*}
\begin{equation*} or, \, f(x) = 4 \frac{x^4}{4}- 3\frac{x^{-3}}{-3})+C = x^4+x^{-3}+C \end{equation*}
\begin{equation*} or, \, f(2) = 0 = 2^4+2^{-3}+C = 16+\frac{1}{8}+C \Rightarrow \, C = -16-\frac{1}{8} = -\frac{129}{8} \end{equation*}
\begin{equation*} \therefore \, f(x) = x^4+x^{-3}-\frac{129}{8} \end{equation*}
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Example 4.1.7.
Find the integration of
  1. \begin{equation*} \int \tan^2 x \, dx \end{equation*}
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  2. \begin{equation*} \int \frac{1}{\sec x-\tan x} \, dx \end{equation*}
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  3. \begin{equation*} \int \frac{1}{1+\cos 2x} \, dx \end{equation*}
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  4. \begin{equation*} \int \frac{1}{1+\sin x} \, dx \end{equation*}
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  5. \begin{equation*} \int \frac{\sin^6 x+\cos^6 x}{\sin^2 x\cos^2 x} \, dx \end{equation*}
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  6. \begin{equation*} \int \frac{\cos x-\cos 2x}{1-\cos x} \, dx \end{equation*}
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Solution.
  1. \begin{equation*} \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C \end{equation*}
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  2. \begin{equation*} \int \frac{1}{\sec x-\tan x} \, dx = \int \frac{\sec x+\tan x}{(\sec x-\tan x)(\sec x+\tan x)} \, dx \end{equation*}
    \begin{equation*} = \int \frac{\sec x+\tan x}{1} \, dx = \int \sec x\, dx +\int \tan x\, dx \end{equation*}
    \begin{equation*} = \log |\sec x +\tan x| -\log |\sec x| +C \end{equation*}
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  3. \begin{equation*} \int \frac{1}{1+\cos 2x} \, dx = \int \frac{1}{2\cos^2 x} \, dx = \int \frac{\sec^2 x}{2} \, dx = \frac{1}{2}\tan x +C \end{equation*}
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  4. \begin{equation*} \int \frac{1}{1+\sin x} \, dx = \int \frac{1-\sin x}{(1+\sin x)(1-\sin x)} \, dx = \int \frac{1-\sin x}{\cos^2 x} \, dx \end{equation*}
    \begin{equation*} = \int \sec^2 x\, dx -\int \sec x\tan x\, dx = \tan x -\sec x +C \end{equation*}
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  5. \begin{equation*} \int \frac{\sin^6 x+\cos^6 x}{\sin^2 x\cos^2 x} \, dx = \int \frac{(\sin^2 x)^3+(\cos^2 x)^3}{\sin^2 x\cos^2 x} \, dx \end{equation*}
    \begin{equation*} = \int \frac{(\sin^2 x +\cos^2 x)((\sin^4 x)-(\sin^2 x)(\cos^2 x)+(\cos^4 x))}{\sin^2 x\cos^2 x} \, dx \end{equation*}
    \begin{equation*} = \int \frac{1-3\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x} \, dx = \int \csc^2 x\, dx -3\int \sec^2 x\, dx \end{equation*}
    \begin{equation*} = -\cot x -3\tan x +C \end{equation*}
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  6. \begin{equation*} \int \frac{\cos x-\cos 2x}{1-\cos x} \, dx = \int \frac{\cos x-(2\cos^2 x -1)}{1-\cos x} \, dx \end{equation*}
    \begin{equation*} = \int \frac{(2\cos x+1)(1-\cos x)}{1-\cos x} \, dx = \int (1+2\cos x) \, dx \end{equation*}
    \begin{equation*} = x +2\sin x +C \end{equation*}
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Example 4.1.8.
Find the integration of
  1. \begin{equation*} \int \frac{2^x+3^x}{5^x}\, dx \end{equation*}
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  2. \begin{equation*} \int \frac{x^4}{x^2+1} \, dx \end{equation*}
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  3. \begin{equation*} \int \frac{x^6-1}{x^2+1} \, dx \end{equation*}
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Solution.
  1. \begin{equation*} \int \frac{2^x+3^x}{5^x}\, dx = \int \left(\left(\frac{2}{5}\right)^x +\left(\frac{3}{5}\right)^x\right) \, dx \end{equation*}
    \begin{equation*} = \int \left(\frac{2}{5}\right)^x \, dx +\int \left(\frac{3}{5}\right)^x \, dx \end{equation*}
    \begin{equation*} = \frac{\left(\frac{2}{5}\right)^x}{\ln \left(\frac{2}{5}\right)} +\frac{\left(\frac{3}{5}\right)^x}{\ln \left(\frac{3}{5}\right)} +C \end{equation*}
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  2. \begin{equation*} \int \frac{x^4}{x^2+1} \, dx = \int (x^2-1) \, dx +\int \frac{1}{x^2+1} \, dx \end{equation*}
    \begin{equation*} = \frac{x^3}{3}-x+\tan^{-1} x +C \end{equation*}
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  3. \begin{equation*} \int \frac{x^6-1}{x^2+1} \, dx = \int \frac{x^6+1-2}{x^2+1} \, dx \end{equation*}
    \begin{equation*} =\int (x^4-x^2+1) \, dx -2 \int \frac{1}{x^2+1} \, dx \end{equation*}
    \begin{equation*} = \frac{x^5}{5}-\frac{x^3}{3}+x-2\tan^{-1} x +C \end{equation*}
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Subsubsection 4.1.1.2 Integration by Substitution Method:

Example 4.1.9.
Find the antiderivative (or integral) of the following functions by substitution method.
  1. \begin{equation*} \int \sin (3x+1)\, dx \end{equation*}
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  2. \begin{equation*} \int \frac{1}{5-2x} \, dx \end{equation*}
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  3. \begin{equation*} \int \left((3x-2)^5+ \frac{1}{(5x+4)^3}\right) \, dx \end{equation*}
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  4. \begin{equation*} \int \frac{1}{\sqrt{2x+3}} \, dx \end{equation*}
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  5. \begin{equation*} \int e^{5x-2} \, dx \end{equation*}
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Solution.
  1. Let
    \begin{equation*} 3x+1=t \end{equation*}
    \begin{equation*} or, \, 3\, dx = dt \end{equation*}
    \begin{equation*} \therefore \, \int \sin (3x+1)\, dx = \int \sin t \cdot \frac{dt}{3} = -\frac{\cos t}{3}+C = -\frac{\cos (3x+1)}{3}+C \end{equation*}
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  2. Let
    \begin{equation*} 5-2x=t \end{equation*}
    \begin{equation*} or, \, -2\, dx = dt \end{equation*}
    \begin{equation*} \therefore \, \int \frac{1}{5-2x} \, dx = -\frac{1}{2}\int \frac{dt}{t} = -\frac{1}{2}\log |t| +C = -\frac{1}{2}\log |5-2x| +C \end{equation*}
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  3. Let
    \begin{equation*} 3x-2=t \, \text{ and } \, 5x+4=u \end{equation*}
    \begin{equation*} or, \, 3\, dx = dt \, \text{ and } \, 5\, dx = du \end{equation*}
    \begin{equation*} \therefore \, \int \left((3x-2)^5+ \frac{1}{(5x+4)^3}\right) \, dx = \frac{1}{3}\int t^5\, dt +\frac{1}{5}\int \left(\frac{3}{u}\right)^3\, du \end{equation*}
    \begin{equation*} = \frac{1}{3}\frac{t^6}{6} +\frac{1}{5} \frac{u^{-2}}{-2}+C= \frac{(3x-2)^6}{18} -\frac{1}{10(5x+4)^2}+C \end{equation*}
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  4. Let \(2x+3=t\) \(or, \, 2\, dx = dt\)
    \begin{equation*} \therefore \, \int \frac{1}{\sqrt{2x+3}} \, dx = \int \frac{1}{\sqrt{t}} \cdot \frac{dt}{2} = \int \frac{t^{-\frac{1}{2}}}{2} \, dt = t^{\frac{1}{2}} +C = \sqrt{2x+3}+C \end{equation*}
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  5. Let
    \begin{equation*} 5x-2=t \end{equation*}
    \begin{equation*} or, \, 5\, dx = dt \end{equation*}
    \begin{equation*} \therefore \, \int e^{5x-2} \, dx = \int e^t \cdot \frac{dt}{5} = \frac{e^t}{5}+C = \frac{e^{5x-2}}{5}+C \end{equation*}
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Example 4.1.10.
\begin{equation*} \int \frac{\sin (\tan^{-1} x)}{(1+x^2)}\, dx \end{equation*}
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Solution.
In this kind of function we need to find the derivative of some function which is already in the problem. For example: Let
\begin{equation*} \tan^{-1} x = t \end{equation*}
differentiate w.r.t."\(t\)", we get -
\begin{equation*} \frac{d}{dx}(\tan^{-1}x) = \frac{d t}{dx} \end{equation*}
\begin{equation*} or, \, \left(\frac{1}{1+x^2}\right)= \frac{d t}{dx} \end{equation*}
\begin{equation*} or, \, \left(\frac{1}{1+x^2}\right)\, dx = dt \end{equation*}
\begin{equation*} Now, \, \int \frac{\sin (\tan^{-1} x)}{(1+x^2)}\, dx = \int \sin t\, dt = -\cos t+C \end{equation*}
\begin{equation*} = -\cos(\tan^{-1}x)+C \end{equation*}
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Example 4.1.11.
Integrate the functions:
  1. \begin{equation*} \frac{2x}{1+x^2} \end{equation*}
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  2. \begin{equation*} \frac{(\log x)^2}{x} \end{equation*}
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  3. \begin{equation*} \frac{1}{x+x\log x} \end{equation*}
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Solution.
  1. Let
    \begin{equation*} 1+x^2 = t \end{equation*}
    \begin{equation*} or, \, 0+2x\, dx = dt \end{equation*}
    \begin{equation*} Now, \, \int \frac{2x}{1+x^2} \, dx = \int \frac{dt}{t} = \log t+C \end{equation*}
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  2. put
    \begin{equation*} \log x=t \end{equation*}
    \begin{equation*} or, \, \frac{1}{x}\, dx =dt \end{equation*}
    \begin{equation*} \therefore \, \int \frac{(\log x)^2}{x}\, dx = \int t^2\, dt = \frac{t^3}{3}=\frac{1}{3}(\log x)^3+C \end{equation*}
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  3. \begin{equation*} \frac{1}{x+x\log x} = \frac{1}{x[1+\log x]} \end{equation*}
    \begin{equation*} \int \frac{1}{x[1+\log x]} \, dx \end{equation*}
    put \(1+\log x =t\)
    \begin{equation*} or, \, \frac{1}{x} \, dx =dt \end{equation*}
    \begin{equation*} or, \, \int \frac{1}{x[1+\log x]} \, dx = \int \frac{dt}{t}= \log |t|+C =\log(1+\log x) +C \end{equation*}
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Subsubsection 4.1.1.3 Integration using Partial Fractions:

The method of partial fractions is a technique used to break down a rational function into simpler fractions that can be easily integrated. This method is particularly useful when the degree of the numerator is less than the degree of the denominator. The process involves expressing the rational function as a sum of simpler fractions, which can then be integrated separately.
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\begin{equation*} \begin{array}{c|c} \text{ Factor in Denominator } & \text{ Term in Partial Fraction} \\ \hline (ax+b) & \frac{A}{ax+b} \\ (ax+b)^n & \frac{A_1}{(ax+b)} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_n}{(ax+b)^n} \\ (ax^2+bx+c) & \frac{Ax+b}{ax^2+bx+c} \\ (ax^2+bx+c)^n & \frac{A_1 x+B_1}{(ax^2+bx+c)} + \frac{A_2 x+B_2}{(ax^2+bx+c)^2}+ \cdots + \frac{A_n x+B_n}{(ax^2+bx+c)^n} \\ \end{array} \text {where } n=1,2,3,\cdots. \end{equation*}
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Example 4.1.12.
  1. \begin{equation*} \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \end{equation*}
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  2. \begin{equation*} \frac{1}{(x+3)^2} = \frac{A}{x+3} + \frac{B}{(x+3)^2} \end{equation*}
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  3. \begin{equation*} \frac{1}{x^2+4} = \frac{Ax+B}{x^2+4} \end{equation*}
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  4. \begin{equation*} \frac{1}{(x^2+4)^2} = \frac{A_1 x+B_1}{x^2+4} + \frac{A_2 x+B_2}{(x^2+4)^2} \end{equation*}
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  5. \begin{equation*} \frac{1}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \end{equation*}
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  6. \begin{equation*} \frac{1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \end{equation*}
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  7. \begin{equation*} \frac{1}{3x^3+4x^2-4x} = \frac{1}{x(3x^2+4x-4)} = \frac{1}{x(3x-2)(x+2)} = \frac{A}{x} + \frac{B}{3x-2} + \frac{C}{x+2} \end{equation*}
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Example 4.1.13.
  1. \begin{equation*} \int \frac{3x+11}{x^2-x-6} \end{equation*}
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  2. \begin{equation*} \int \frac{5x-2}{(x+3)^2} \end{equation*}
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Solution.
  1. \begin{equation*} \int \frac{3x+11}{x^2-x-6} \end{equation*}
    • Step 1: Factor the denominator \(x^2 - x - 6\) as: \((x-3)(x+2)\)
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    • Step 2: Partial fraction decomposition of the function \(\frac{3x+11}{(x-3)(x+2)}\) is: \(\frac{A}{x-3} + \frac{B}{x+2}\)
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    • Step 3: Find the values of A and B by multiplying both sides of the equation by the denominator \((x-3)(x+2)\) and then equating the coefficients of the powers of x on both sides of the equation. We get: \(A(x+2) + B(x-3) = 3x + 11\)
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    • Expand:
      \begin{equation*} 3x+11=Ax+2A+Bx-3B=(A+B)x+(2A-3B) \end{equation*}
      Equating the coefficients of x and the constant term, we get: \(A+B=3\) and \(2A-3B=11\) Solving these equations, we get: \(A=4\) and \(B=-1\)
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    • Step 4: Now, we can write the function as: \(\frac{3x+11}{(x-3)(x+2)} = \frac{4}{x-3} - \frac{1}{x+2}\)
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    • Step 5: Now, we can integrate the function as:
      \begin{equation*} \int \frac{3x+11}{x^2-x-6} \, dx = \int \left(\frac{4}{x-3} - \frac{1}{x+2}\right) \, dx \end{equation*}
      \begin{equation*} = 4\int \frac{1}{x-3} \, dx - \int \frac{1}{x+2} \, dx = 4\log |x-3| - \log |x+2| +C \end{equation*}
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  2. \begin{equation*} \int \frac{5x-2}{(x+3)^2} \end{equation*}
    • Step 1: Partial fraction decomposition of the function \(\frac{5x-2}{(x+3)^2}\) is: \(\frac{A}{x+3} + \frac{B}{(x+3)^2}\)
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    • Step 2: Find the values of A and B by multiplying both sides of the equation by the denominator \((x+3)^2\) and then equating the coefficients of the powers of x on both sides of the equation. We get: \(A(x+3) + B = 5x - 2\)
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    • Step 3: Expand:
      \begin{equation*} 5x-2=Ax+3A+B \end{equation*}
      Equating the coefficients of x and the constant term, we get: \(A=5\) and \(3A+B=-2\) Solving these equations, we get: \(A=5\) and \(B=-17\)
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    • Step 4: Now, we can write the function as: \(\frac{5x-2}{(x+3)^2} = \frac{5}{x+3} - \frac{17}{(x+3)^2}\)
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    • Step 5: Now, we can integrate the function as:
      \begin{equation*} \int \frac{5x-2}{(x+3)^2} \, dx = \int \left(\frac{5}{x+3} - \frac{17}{(x+3)^2}\right) \, dx \end{equation*}
      \begin{equation*} = 5\int \frac{1}{x+3} \, dx - 17\int \frac{1}{(x+3)^2} \, dx = 5\log |x+3| + \frac{17}{x+3} +C \end{equation*}
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Subsubsection 4.1.1.4 Integration by Parts:

The integration by parts is a method of integration which is used to integrate the product of two functions. The formula for integration by parts is given by:
\begin{equation*} \int u \, v\, dx = u\, \int v \, dx - \int \{\frac{du}{dx}\int v\, dx \}\, dx \end{equation*}
where \(u\) and \(v\) are two functions of \(x\text{.}\)
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A commonly used mnemonics for choosing the first function and the second function in integration by parts is the ILATE rule (choose u in this priority order): Inverse trigonometric, Logarithmic, Algebraic (polynomials), Trigonometric, Exponential.
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Example 4.1.14.
  1. \begin{equation*} \int x \sin x \, dx \end{equation*}
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  2. \begin{equation*} \int x^2 e^x \, dx \end{equation*}
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  3. \begin{equation*} \int x \log x \, dx \end{equation*}
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  4. \begin{equation*} \int e^x \sin x \, dx \end{equation*}
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  5. \begin{equation*} \int x^2 \cos x \, dx \end{equation*}
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Solution.
  1. Let \(u=x\) and \(dv=\sin x \, dx\)
    \begin{equation*} or, \, du=dx \, \text{ and } \, v=-\cos x \end{equation*}
    \begin{equation*} \therefore \, \int x \sin x \, dx = -x\cos x +\int \cos x \, dx = -x\cos x +\sin x +C \end{equation*}
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  2. Let \(u=x^2\) and \(dv=e^x \, dx\)
    \begin{equation*} or, \, du=2x\, dx \, \text{ and } \, v=e^x \end{equation*}
    \begin{equation*} \therefore \, \int x^2 e^x \, dx = x^2 e^x - 2\int xe^x\, dx \end{equation*}
    Now, we need to apply integration by parts again to evaluate the integral \(\int xe^x\, dx\text{.}\) Let \(u=x\) and \(dv=e^x\, dx\)
    \begin{equation*} or, \, du=dx \, \text{ and } \, v=e^x \end{equation*}
    \begin{equation*} \therefore \, 2\int xe^x\, dx = 2(xe^x -\int e^x\, dx) = 2xe^x - 2e^x +C \end{equation*}
    Now we can write the final answer as:
    \begin{equation*} \int x^2 e^x \, dx = x^2 e^x - (2xe^x - 2e^x) +C = (x^2 - 2x + 2)e^x +C \end{equation*}
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  3. Let \(u=\log x\) and \(dv=x\, dx\)
    \begin{equation*} or, \, du=\frac{1}{x}\, dx \, \text{ and } \, v=\frac{x^2}{2} \end{equation*}
    \begin{equation*} \therefore \, \int x \log x \, dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \log x - \frac{1}{2}\int x \, dx \end{equation*}
    \begin{equation*} = \frac{x^2}{2} \log x - \frac{x^2}{4} +C \end{equation*}
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  4. Let \(u=\sin x\) and \(dv=e^x \, dx\)
    \begin{equation*} or, \, du=\cos x\, dx \, \text{ and } \, v=e^x \end{equation*}
    \begin{equation*} \therefore \, \int e^x \sin x \, dx = -e^x \sin x + \int e^x \cos x \, dx \end{equation*}
    Now, we need to apply integration by parts again to evaluate the integral \(\int e^x \cos x \, dx\text{.}\) Let \(u=\cos x\) and \(dv=e^x\, dx\)
    \begin{equation*} or, \, du=-\sin x\, dx \, \text{ and } \, v=e^x \end{equation*}
    \begin{equation*} \therefore \, \int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx \end{equation*}
    Now we can write the final answer as:
    \begin{equation*} \int e^x \sin x \, dx = -e^x \sin x + e^x \cos x + \int e^x \sin x \, dx \end{equation*}
    \begin{equation*} \Rightarrow \, 2\int e^x \sin x \, dx = e^x (\cos x - \sin x) +C \end{equation*}
    \begin{equation*} \therefore \, \int e^x \sin x \, dx = \frac{e^x (\cos x - \sin x)}{2} +C \end{equation*}
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  5. Let \(u=x^2\) and \(dv=\cos x \, dx\)
    \begin{equation*} or, \, du=2x\, dx \, \text{ and } \, v=\sin x \end{equation*}
    \begin{equation*} \therefore \, \int x^2 \cos x \, dx = x^2 \sin x - 2\int x \sin x \, dx \end{equation*}
    Now, we need to apply integration by parts again to evaluate the integral \(\int x \sin x \, dx\text{.}\) Let \(u=x\) and \(dv=\sin x\, dx\)
    \begin{equation*} or, \, du=dx \, \text{ and } \, v=-\cos x \end{equation*}
    \begin{equation*} \therefore \, \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x +C \end{equation*}
    Now we can write the final answer as:
    \begin{equation*} \int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) +C \end{equation*}
    \begin{equation*} = x^2 \sin x + 2x \cos x - 2\sin x +C \end{equation*}
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