Question

What is `int(lnx^3)/x^4 dx`?

Answer 1

`-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+C`

Explanation:

Use integration-by-parts with `u=ln(x^3)=3ln(x)` and `dv=1/(x^4)dx=x^(-4)dx` to get `du=3/x` and `v=x^(-3)/(-3)`.

Then, since `int\ u\ dv=uv-int\ v\ du`, we get

`int\ (ln(x^3))/(x^4)\ dx= int\ (3ln(x))/(x^4)\ dx`

`=-(ln(x^3))/(3x^3)-int\ -x^(-4)\ dx=-(ln(x^3))/(3x^3)+int\ x^(-4)\ dx`

`=-(ln(x^3))/(3x^3)+(x^(-3))/(-3)+C`

`=-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+C`