Question

How do you evaluate the integral of `int x^3 lnx dx`?

Answer 1

`intx^3lnxdx=(x^4(4lnx-1))/16+C`

Explanation:

Use integration by parts, which states that:

`intudv=uv-intvdu`

So, for `intx^3lnxdx`, let `u=lnx` and `dv=x^3dx`.

These imply that `du=1/xdx` and `v=x^4/4` (obtain these by differentiating `u` and integrating `dv`, respectively).

Plugging these into the integration by parts formula, this yields:

`intx^3lnxdx=lnx(x^4/4)-int(x^4/4)(1/x)dx`

`=(x^4lnx)/4-1/4intx^3dx`

`=(x^4lnx)/4-x^4/16+C`

`=(x^4(4lnx-1))/16+C`