Question

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

Answer 1

`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`