StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 34

Expert-verifiedFound in: Page 464

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 $x-{\mathrm{tan}}^{-1}x+C.$

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

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,*

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

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

94% of StudySmarter users get better grades.

Sign up for free