Question

How do you find the indefinite integral of `int (x+1/x)^2dx`?

Answer 1

`int(x+1/x)^2dx=x^3/3+2x-1/x+C`

Explanation:

Note that:

`(x+1/x)^2=(x+1/x)(x+1/x)`

`color(white)((x+1/x)^2)=x^2+2x(1/x)+(1/x)^2`

`color(white)((x+1/x)^2)=x^2+2+x^-2`

So:

`I=int(x+1/x)^2dx=intx^2dx+2intdx+intx^-2dx`

Using `intdx=x` and `intx^ndx=x^(n+1)/(n+1)` we obtain:

`I=x^3/3+2x+x^-1/(-1)+C`

`I=x^3/3+2x-1/x+C`