Question

Evaluate the following integral:

`int (x + 1)^2 log x  "d"x`

Answer 1

`int (x + 1)^2 log x  "d"x`

We use integration by parts method.

Let u = log x

⇒ du = `1/x  "d"x`

dv = `(x + 1)^2  "d"x`

So v = `(x + 1)^3/3`

We have `int (x + 1)^2 log x  "d"x = (x + 1)^3/3 log x - int (x + 1)^3/3 (1/x)  "d"x`

= `(x + 1)^3/3 log x - 1/3 int ((x^3 + 3x^2 + 3x + 1))/x  "d"x`

= `(x + 1)^3/3 log x - 1/3 int (x^2 + 3x + 3 + 1/x)  "d"x`

= `1/3[(x + 1)^3 log x - x^3/3 - (3x^2)/2 - 3x - log |x|] + "c"`