Question

What is the antiderivative of `x/(7x^2+3)^5`?

Answer 1

`-1/(56(7x^2+3)^4)+C`

Explanation:

We wish to find:

`intx/(7x^2+3)^5dx`

This is a prime case for substitution. Let `u=7x^2+3` so that `du=14xdx`.

Note that we already have `xdx` in the numerator, so we need to multiply the integrand by `14` to obtain `du`. Balance this by multiplying the exterior of the integral by `1/14`.

`=1/14int(14x)/(7x^2+3)^5dx`

Now substitute in `u` and `du`.

`=1/14int(du)/u^5=1/14intu^-5du`

Integrating `u^-5` results in `u^(-5+1)/(-5+1)+C`:

`=1/14(u^-4/(-4))+C=-1/(56u^4)+C=-1/(56(7x^2+3)^4)+C`