• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q42E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 335
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

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 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: \(I = \int x ta{n^{ - 1}}xdx = \frac{{{x^2} + 1}}{2}ta{n^{ - 1}}x - \frac{x}{2} + c\).

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.