Question

What is the integral of `x^3/(x^2+1)`?

Answer 1

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

Explanation:

We have the integral:

`intx^3/(x^2+1)dx`

We will use substitution: let `u=x^2+1`, implying that `du=2xdx`.

Rearrange the integral, including making `2xdx` present:

`intx^3/(x^2+1)dx=int(x^2*x)/(x^2+1)dx=1/2int(x^2*2x)/(x^2+1)dx`

Make the following substitutions into the integral:

`{(x^2+1=u),(x^2=u-1),(2xdx=du):}`

We obtain:

`1/2int(x^2*2x)/(x^2+1)dx=1/2int(u-1)/udu`

Splitting the integral up through subtraction:

`1/2int(u-1)/udu=1/2int1du-1/2int1/udu`

These are common integrals:

`=1/2u-1/2ln(absu)+C`

Since `u=x^2+1`:

`=1/2(x^2+1)-1/2ln(abs(x^2+1))+C`

The absolute value bars are not needed since `x^2+1>0` for all Real values of `x`. Also note that `1/2(x^2+1)=1/2x^2+1/2`, so the `1/2` gets absorbed into the constant of integration `C`:

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

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