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Q. 34

Expert-verified
Found in: Page 464

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Solve $\int \frac{{x}^{2}}{1+{x}^{2}}dx$ the following two ways:(a) with the substitution $u={\mathrm{tan}}^{-1}x;$(b) with the trigonometric substitution x = tan u.

Part (a) The solution of the given integral is $x-{\mathrm{tan}}^{-1}x+C.$

Part (b) The solution of the given integral is $x-{\mathrm{tan}}^{-1}x+C.$

See the step by step solution

## Part (a) Step 1. Given Information.

The given integral is $\int \frac{{x}^{2}}{1+{x}^{2}}dx.$

## Part (a) Step 2. Solve.

We have to solve the integral with the substitution $u={\mathrm{tan}}^{-1}x,$ so the derivative is $du=\frac{1}{1+{x}^{2}}dx.$

Let's solve the integral by substituting u,

$\int \frac{{x}^{2}}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=\int \frac{{\left(\mathrm{tan}u\right)}^{2}}{1+{\left(\mathrm{tan}u\right)}^{2}}\left(1+{\left(\mathrm{tan}u\right)}^{2}\right)du\phantom{\rule{0ex}{0ex}}=\int {\mathrm{tan}}^{2}udu\phantom{\rule{0ex}{0ex}}=\int -1+se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=-\int 1du+\int se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=-u+\mathrm{tan}u+C\phantom{\rule{0ex}{0ex}}\mathrm{Substitute}\mathrm{back}u\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}x+x+C\phantom{\rule{0ex}{0ex}}=x-{\mathrm{tan}}^{-1}x+C$

## Part (b) Step 1. Solve.

We have to solve the integral with the substitution $x=\mathrm{tan}u,$ so the derivative is $dx=se{c}^{2}udu.$

Let's solve the integral by substituting x,

role="math" localid="1648810032401" $\int \frac{{x}^{2}}{1+{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=\int \frac{{\mathrm{tan}}^{2}u}{1+{\mathrm{tan}}^{2}u}se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=\int \frac{{\mathrm{tan}}^{2}u}{se{c}^{2}u}se{c}^{2}udu\left[1+{\mathrm{tan}}^{2}u=se{c}^{2}u\right]\phantom{\rule{0ex}{0ex}}=\int {\mathrm{tan}}^{2}udu\phantom{\rule{0ex}{0ex}}=\int -1+se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=-\int 1du+\int se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=-u+\mathrm{tan}u+C\phantom{\rule{0ex}{0ex}}\mathrm{Substitute}\mathrm{back}u\phantom{\rule{0ex}{0ex}}=-{\mathrm{tan}}^{-1}x+x+C\phantom{\rule{0ex}{0ex}}=x-{\mathrm{tan}}^{-1}x+C$