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Section 5.6 Antiderivatives (Integration)

Subsection 5.6.1 Inspection and Substitution

Checkpoint 5.6.1.

Use the inspection method to find the general antiderivative: \(\int (6x^2 - 4x + 5) \, dx\)

Checkpoint 5.6.2.

Checkpoint 5.6.3.

By inspection, find the antiderivative of \(f(x) = \sec^2(x) + \sin(x)\text{.}\)

Checkpoint 5.6.4.

Checkpoint 5.6.5.

Evaluate \(\int \frac{(\ln x)^2}{x} \, dx\) using \(u = \ln x\text{.}\)

Checkpoint 5.6.6.

Checkpoint 5.6.7.

Checkpoint 5.6.8.

Checkpoint 5.6.9.

Checkpoint 5.6.10.

Subsection 5.6.2 Integration by Parts and Partial Fractions

Checkpoint 5.6.11.

Evaluate \(\int \frac{1}{(x-1)(x+2)} \, dx\)
  • \(\frac{1}{3}\ln|x-1| - \frac{1}{3}\ln|x+2| + C\)
  • \(\ln|x-1| + \ln|x+2| + C\)
  • \(\frac{1}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C\)

Checkpoint 5.6.12.

Checkpoint 5.6.13.

Evaluate \(\int \frac{1}{x^2-4} \, dx\)
  • \(\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C\)
  • \(\frac{1}{2}\ln\left|\frac{x-2}{x+2}\right| + C\)
  • \(\ln|x^2-4| + C\)

Checkpoint 5.6.14.

Checkpoint 5.6.15.

Checkpoint 5.6.16.

Checkpoint 5.6.17.

Checkpoint 5.6.18.

Checkpoint 5.6.19.

Checkpoint 5.6.20.

Checkpoint 5.6.21.

Use integration by parts to evaluate: \(\int x e^x \, dx\)
Hint.
Correct! Use integration by parts with \(u = x\text{,}\) \(dv = e^x dx\text{.}\)
Incorrect. Remember: \(\int u\,dv = uv - \int v\,du\text{.}\)

Checkpoint 5.6.22.

Checkpoint 5.6.23.

Use partial fraction decomposition to find: \(\int \frac{1}{x^2 - 1} \, dx\)
  • \(\frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C\)
  • \(\ln|x^2 - 1| + C\)
  • \(\frac{1}{2}\ln|x+1| - \frac{1}{2}\ln|x-1| + C\)

Checkpoint 5.6.24.

What is the correct form of the partial fraction decomposition for the integrand \(\frac{5x}{(x-2)(x^2+1)}\text{?}\)
  • \(\frac{A}{x-2} + \frac{Bx+C}{x^2+1}\)
  • \(\frac{A}{x-2} + \frac{B}{x^2+1}\)
  • \(\frac{Ax+B}{x-2} + \frac{Cx+D}{x^2+1}\)

Checkpoint 5.6.25.