Question

Integrate the function: $xtan^{-1}x$

Answer 1

Let $I$ =∫$$xtan^{-1}x\ dx

Taking as first function and x as second function and integrating by parts, we obtain

$I$ = tan-1x∫x dx-∫{($\frac {d}{dx}$tan-1x)∫x dx}dx

$I$= tan-1x ($\frac {x^2}{2}$)-$∫\frac {1}{1+x^2}.\frac {x^2}{2} dx$

$I$= $\frac {x^2tan^{-1}x}{2}$ - \frac 12$$∫\frac {x^2}{1+x^2} dx

$I$= $\frac {x^2tan^{-1}x}{2}$ - \frac 12$$∫(\frac {x^2+1}{1+x^2}-\frac {1}{1+x^2})dx

$I$= $\frac {x^2tan^{-1}x}{2}$ - \frac 12$$∫(1-\frac {1}{1+x^2})dx

$I$= $\frac {x^2tan^{-1}x}{2}$ - \frac 12$$(x-tan^{-1}x)+C

$I$= $\frac {x^2}{2}tan^{-1}x - \frac x2+\frac 12tan^{-1}x+C$