What is $int tan^-1 x dx$ ?

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Answer 1 · 2018-02-20T09:48:50

$I=tan^-1(x)x-1/2ln(x^2+1)+C$

We want to solve

$I=inttan^-1(x)dx$

$intudv=uv-intvdu$

Let $u=tan^-1(x)$ and $dv=1dx$

Then $du=1/(x^2+1)dx$ and $v=x$

$I=tan^-1(x)x-intx/(x^2+1)dx$

Make a substitution $u=x^2+1=>(du)/dx=2x$

$I=tan^-1(x)x-1/2int1/(u)du$

$=tan^-1(x)x-1/2ln(u)+C$

Substitute back $u=x^2+1$

$I=tan^-1(x)x-1/2ln(x^2+1)+C$

What is int tan^-1 x dx int tan^-1 x dx ?