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#