Question

Question #64159

Answer 1

`int(x+1/x)^2dx`

Expand the square:

`=int(x+1/x)(x+1/x)dx`

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

`=int(x^2+2+x^-2)dx`

Integrate these using `intx^ndx=x^(n+1)/(n+1)` and `aintdx=ax`:

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

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

Getting a common denominator:

`=(x^4+6x^2-3)/(3x)+C`