Question

How do you find the integral of `x^3/(1+x^2)`?

Answer 1

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

Explanation:

We'll do integration by substitution. Take:
`u = x^2`
`du = 2x`
Substituting this in, we get:

`1/2\intu/(1+u)du`
Now take:
`t = 1 + u`
`dt = du`

Substitute this in:
`1/2\int(t-1)/tdt = 1/2\int1 - 1/tdt = 1/2(\int1dt - \int 1/tdt)`
`=1/2(t - ln(t)) + C`
Now we need to change back to `x`:
`=1/2(1+u-ln(1+u)) + C`
`=1/2(1+x^2-ln(1+x^2)) + C`