Question

How do you integrate `int x^3 ln x^4 dx` using integration by parts?

Answer 1

`intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C`

Explanation:

First, let's rewrite `ln(x^4)` as `4ln(x),` as `ln(x^a)=aln(x)`

`intx^3ln(x^4)dx=int4x^3ln(x)dx=4intx^3ln(x)dx` (We can factor out `4,` it's just a constant.)

Make the following choices for `u` and `dv:`

`u=ln(x)`

`dv=x^3dx`

Thus, `du` and `v` become:

`du=1/xdx`

`dv=intx^3dx=1/4x^4` (We're not going to include the constant just yet. It can wait until the end.)

Now it becomes apparent why we made the choices we did for `u` and `v.` If we chose `dv=ln(x),` we would have to integrate `ln(x)` for `v,` and would run into our original problem.

Plug in relevant values:

`uv-intvdu=1/4x^4ln(x)-int1/4x^4/xdx`

We can simplify the integral here:

`1/4x^4ln(x)-int1/4x^4/xdx=1/4x^4ln(x)-1/4intx^(cancel(4)3)/cancelx`

Thus, we have

`1/4x^4ln(x)-1/4intx^3dx=1/4x^4ln(x)-(1/4)(1/4)x^4+C=1/4x^4ln(x)-1/16x^4+C`

Let's not forget the `4` we originally factored out...

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

`C` remains unchanged; a constant multiplied by a constant remains a constant.

So,

`intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C`

Answer 2

`int\ x^3ln(x^4)\ "d"x=4x^4ln(x)-x^4+C`

Explanation:

Here's another approach:

Consider substituting first `w=x^4` and `("d"w)/("d"x)=4x^3` into the integral `int\ x^3ln(x^4)\ "d"x`.

This gives `1/4int\ ln(w)\ "d"w`.

Now use integration by parts `int\ u\ "d"v=uv-int\ v\ "d"u`. Use `u=ln(w)` and `"d"v="d"w`. Then find that `"d"u=("d"w)/w` and `v=w`.

Now, we have `1/4int\ ln(w)\ "d"w=1/4(wln(w)-int\ "d"w)`
or simply `1/4wln(w)-1/4w+C`.

Substitute back `w=x^4`.

Our final answer is thus `1/4x^4ln(x^4)-1/4x^4+C=x^4ln(x)-1/4x^4+C`.