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

Question

Cameron Alexander · Accepted Answer

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

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):}#

The original integral therefore 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))#

Consequently:

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

Ezra Adams · Answer

undefined The integral of $ \frac{\arctan(x)}{x^2} $ is not expressible in terms of elementary functions. It cannot be represented using standard functions such as polynomials, exponentials, logarithms, trigonometric functions, and their inverses. However, it can be expressed in terms of special functions, such as the dilogarithm function.