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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int x\: tan^{-1}x\: dx$

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Toolbox:
  • 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$
$v=\large\frac{x^2}{2}$
$dv=xdx$
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}$$dx$
$\qquad\qquad\qquad=\large\frac{x^2\tan^{-1}x}{2}-\frac{x}{2}+\large\frac{\tan^{-1}x}{2}$$+c$
$\qquad\qquad\qquad=\large\frac{1}{2}$$\big[x^2\tan^{-1}x-x+\tan^{-1}x\big]+c$
answered Sep 30, 2013 by sreemathi.v
 
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