Question

How do you integrate `(ln x)^2 dx`?

Answer 1

Use integration by parts twice to find that

`intln^2(x)dx=xln^2(x)-2xln(x)+2x+C`

Explanation:

We will proceed using integration by parts :

Integration by parts (i):

Let `u = ln^2(x)` and `dv = dx`
Then `du = 2ln(x)/xdx` and `v = x`

Applying the integration by parts formula `intudv = uv - intvdu`

`intln^2(x)dx = xln^2(x) - int(2xln(x))/xdx`

`=xln^2(x)-2intln(x)dx" (1)"`

Integration by parts (ii):

Focusing on the remaining integral, let `u = ln(x)` and `dv = dx`
Then `du = 1/xdx` and `v = x`

Applying the formula:

`intln(x)dx = xln(x) - intx/xdx`

`=xln(x) - intdx`

`=xln(x) - x + C`

Substituting this back into `"(1)"`:

`intln^2(x)dx = xln^2(x)-2(xln(x)-x)+C`

`=xln^2(x)-2xln(x)+2x+C`