Question

How do you integrate `int ln x^3 dx` using integration by parts?

Answer 1

`intln(x^3)dx=3xlnx-3x+C`

Explanation:

First, simplify the function using the rule `ln(a^b)=b*lna`; this gives the simplified integral

`intln(x^3)dx=int3lnxdx=3intlnxdx`

The question then becomes, how do we integrate `intlnxdx` by parts?

Integration by parts takes the form

`intudv=uv-intvdu`

So, we have to determine what to let `u` and `dv` equal within `intlnxdx`. Our best bet is to let `u=lnx` and simply have `dv=1dx`, since there's not much else we can do.

Find the values of `du` and `v`, by differentiating and integrating respectively.

`ul(u=lnx)" "=>" "(du)/dx=1/x" "=>" "ul(du=1/xdx)`

`ul(dv=1dx)" "=>" "intdv=intdx" "=>" "ul(v=x`

Plugging these into the integration by parts formula, not forgetting the `3` tagging along multiplicatively, we have:

`3intlnxdx=3(xlnx-intx(1/x)dx)`

`=3xlnx-3intdx`

`=3xlnx-3x+C`