Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Evaluate : $ \int x\: tan^{-1}x\: dx$

Can you answer this question?

1 Answer

0 votes
  • When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts $\int udv=uv-\int vdu$
  • $\;\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2}.$
  • Method of substitution :If $f(x)=t$,then $f'(x)dx=dt$.
  • Hence $\int f(x)=\int t.dt+c.$
Step 1:
$I=\int x\tan^{-1}xdx$
This is of the form $\int udv$.This can be solved by $uv-\int vdu$
Let $u=\tan^{-1}x$
$\Rightarrow du=\large\frac{1}{1+x^2}$$dx$
Step 2:
$\therefore \int x\tan^{-1}xdx=(\tan^{-1}x)(\large\frac{x^2}{2})-\int \large\frac{x^2}{2}\frac{1}{1+x^2}$$dx$
$\qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int \frac{x^2}{1+x^2}$$dx$
$\qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int \frac{x^2+1-1}{1+x^2}$$dx$
$\qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int$$ dx-\int \large\frac{1}{1+x^2}$$dx$
answered Sep 30, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App