Question

How do solve `∫x tan^-1x dx`, given that `d/dx tan^-1x = 1/(1+x^2)` ?

Answer 1

`((x^2+1)tan^-1(x)-x)/2+C`

Explanation:

`I=intxtan^-1(x)dx`

We will use integration by parts. Integration by parts takes the form `intudv=uv-intvdu`.

Since we cannot integrate `tan^-1(x)` well, let `u=tan^-1(x)` (notice you've also been provided with the derivative of `tan^-1(x)` in the question). Thus, let `dv` be the remaining portion: `dv=xdx`.

Differentiating `u` and integrating `dv`:

• `u=tan^-1(x)" "=>" "du=1/(1+x^2)dx`
• `dv=xdx" "=>" "v=x^2/2`

Then:

`I=uv-intvdu=x^2/2tan^-1(x)-intx^2/2(1/(1+x^2))dx`

Simplifying slightly:

`I=x^2/2tan^-1(x)-1/2intx^2/(1+x^2)dx`

Rewrite the integrand as follows, or perform polynomial long division on `x^2/(1+x^2)`:

`I=x^2/2tan^-1(x)-1/2int(1+x^2-1)/(1+x^2)dx`

`I=x^2/2tan^-1(x)-1/2int(1+x^2)/(1+x^2)dx-1/2int1/(1+x^2)dx`

`I=x^2/2tan^-1(x)-1/2intdx-1/2int(-1)/(1+x^2)dx`

Note that `intdx=x`. Furthermore, since `d/dxtan^-1(x)=1/(1+x^2)`, we see that `int1/(1+x^2)dx=tan^-1(x)`.

`I=x^2/2tan^-1(x)-1/2x+1/2tan^-1(x)`

Which we may write as:

`I=(x^2tan^-1(x)-x+tan^-1(x))/2`

Rearranging and adding the constant of integration:

`I=((x^2+1)tan^-1(x)-x)/2+C`