Question

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

Answer 1

`int3xln(x^3)dx=(9x^2lnx)/2-9/4x^2+C`

Explanation:

We're going to want to get rid of the `x^3`. Recalling that `ln(x^a)=alnx, ln(x^3)=3lnx,` and we get

`3(3)intxlnxdx=9intxlnxdx` factoring all constants outside.

Now, we'll make the following selections for Integration by Parts:

`u=lnx`

`du=x^-1dx`

`dv=xdx`

`v=intxdx=1/2x^2`

So, applying the formula, we get

`uv-intvdu=(x^2lnx)/2-1/2intx^-1x^2dx`

`=(x^2lnx)/2-1/2intxdx=(x^2lnx)/2-1/4x^2+C`

Recall that we need to multiply through by the `9` we factored out:

`9[(x^2lnx)/2-1/4x^2+C]=(9x^2lnx)/2-9/4x^2+C`

Thus,

`int3xln(x^3)dx=(9x^2lnx)/2-9/4x^2+C`

Answer 2

`int 3xlnx^3dx = (9x^2)/4 (2lnx-1)+C`

Explanation:

Using the properties of logarithms:

`ln x^3 = 3lnx`

so:

`int 3xlnx^3dx = 9 int xlnxdx`

integrate now by parts:

`int 3xlnx^3dx = 9 int lnx d(x^2/2)`

`int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x^2 d(lnx)`

`int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x^2 dx/x`

`int 3xlnx^3dx = (9x^2lnx)/2 - 9/2 int x dx`

`int 3xlnx^3dx = (9x^2lnx)/2 - (9x^2)/4+C`

`int 3xlnx^3dx = (9x^2)/4 (2lnx-1)+C`