Question

What is the indefinite integral of `ln(x^3)/x`?

Answer 1

`int(ln(x^3))/xdx=(3(lnx)^2)/2+C`

Explanation:

We'll start by making life a lot easier and using a property of the natural log to simplfy things:
`ln(x^a)=alnx`

Utilizing this property, `int(ln(x^3))/xdx` becomes `int(3lnx)/xdx`, and furthermore, since 3 is constant, `3intlnx/xdx`.

Now, notice that we have `lnx` and its derivative, `1/x`. This makes the integral a textbook case of a `u`-substitution:
`u=lnx->(du)/dx=1/x->du=1/xdx`

We can rewrite the integral a little to make it easier to follow along:
`3int(lnx)(1/x)dx`

Because `color(red)u=lnx` and `color(blue)(du)=1/xdx`,
`3intcolor(red)(lnx)color(blue)((1/x)dx)=3intcolor(red)ucolor(blue)(du)`

Our new integral is easily evaluated by the reverse power rukle:
`3intudu=3(u^2/2+C)`
`color(white)(XX)=(3u^2)/2+C`

Finally, because `u=lnx`,
`(3u^2)/2+C=(3(lnx)^2)/2+C`