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Found in: Page 335

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral: $$\int {xta{n^{ - 1}}xdx}$$

Apply integration by parts using $$ta{n^{ - 1}}x$$as a first function.

$$ta{n^{ - 1}}x$$ is not subject to specify forward integration so while apply integration by parts, $$ta{n^{ - 1}}x$$ should be taken as the first function.

See the step by step solution

## Step 1: integration by parts

\begin{aligned}{l}\int x ta{n^{ - 1}}xdx &= ta{n^{ - 1}}x\int {xdx - \int {(\frac{d}{{dx}}(ta{n^{ - 1}}x)} } \int {xdx} )dx\\\int x ta{n^{ - 1}}xdx &= \frac{{{x^2}ta{n^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} \\ &= \frac{{{x^2}ta{n^{ - 1}}x}}{2} - \frac{1}{2}{I_1}\end{aligned}

## Step 2: reduction of degree of numerator

\begin{aligned}{l}{I_1} &= \int {\frac{{{x^2}}}{{1 + {x^2}}}dx} &= \int {\frac{{({x^2} + 1) - 1}}{{1 + {x^2}}}dx} &= \int {dx - \int {\frac{{dx}}{{1 + {x^2}}}} } \\ &= x - ta{n^{ - 1}}x\end{aligned}

\begin{aligned}{l}I &= \int x ta{n^{ - 1}}xdx = \frac{{{x^2}ta{n^{ - 1}}x}}{2} - \frac{1}{2}(x - ta{n^{ - 1}}x)\\I &= \int x ta{n^{ - 1}}xdx &= \frac{{{x^2} + 1}}{2}ta{n^{ - 1}}x - \frac{x}{2} + c\end{aligned}

Hence, the solution of the given integral is: $$I = \int x ta{n^{ - 1}}xdx = \frac{{{x^2} + 1}}{2}ta{n^{ - 1}}x - \frac{x}{2} + c$$.