How do you evaluate the integral $int arctanx/x^2$ ?

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Answer 1 · 2017-06-20T15:53:06

$-arctanx/x+lnabsx-1/2lnabs(x^2+1)+C$

$I=intarctanx/x^2dx$

$u=arctanx" "=>" "du=1/(x^2+1)dx$

$dv=1/x^2dx" "=>" "v=-1/x$

Then:

$I=uv-intvdu$

$I=-arctanx/x+intdx/(x(x^2+1))$

Performing partial fractions:

$1/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)$

$1=A(x^2+1)+(Bx+C)x$

$1=x^2(A+B)+x(C)+A$

Comparing coefficients,

${(A+B=0),(C=0),(A=1):}$

So $B=-1$ as well, giving the decomposition:

$1/(x(x^2+1))=1/x+(-x)/(x^2+1)$

So:

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

$I=-arctanx/x+lnabsx-1/2int(2x)/(x^2+1)dx$

$I=-arctanx/x+lnabsx-1/2lnabs(x^2+1)+C$

How do you evaluate the integral int arctanx/x^2int arctanx/x^2?