What is the integral of #arctanx/x^2#?

1 Answer
Sep 13, 2016

#lnabs(x/sqrt(1+x^2))-arctan(x)/x+C#

Explanation:

We have:

#I=intarctan(x)/x^2dx=intx^-2arctan(x)dx#

We will use integration by parts, which takes the form: #intudv=uv-intvdu#

Let:

#{(u=arctan(x)" "=>" "du=dx/(1+x^2)),(dv=x^-2" "=>" "v=-x^-1=-1/x):}#

Thus the original integral equals:

#I=-arctan(x)/x+intdx/(x(1+x^2))#

Letting #J=intdx/(x(1+x^2))#. This can be solved using partial fractions but I prefer trigonometric substitution. Here, let #x=tan(theta)#. Note that #dx=sec^2(theta)d theta#. Thus:

#J=int(sec^2(theta)d theta)/(tan(theta)(1+tan^2(theta)))#

Note that #1+tan^2(theta)=sec^2(theta)#:

#J=int(sec^2(theta)d theta)/(tan(theta)(sec^2(theta)))=intcot(theta)d theta=lnabssin(theta)#

Note that since #x=tan(theta)#, we see that #cot(theta)=1/x#. Note that #csc(theta)=sqrt(cot^2(theta)+1)=sqrt(1/x^2+1)=sqrt(1+x^2)/absx#. Furthermore, we see that #sin(theta)=1/csc(theta)=absx/sqrt(1+x^2)#. Thus:

#J=lnabs(x/sqrt(1+x^2))#

Therefore:

#I=lnabs(x/sqrt(1+x^2))-arctan(x)/x+C#