Question

How do you find `int(x^2-1/x^2+root3x)dx`?

Answer 1

`=x^3/3+1/x+3/4x^(4/3)+C`

Explanation:

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

By rewriting,

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

By the Power Rule for integration, which states that `intx^ndx=x^(n+1)/(n+1)+C`:

`=x^3/3-x^(-1)/(-1)+x^(4/3)/(4/3)+C`

By cleaning up,

`=x^3/3+1/x+3/4x^(4/3)+C`

I hope that this was clear.

Answer 2

`1/3x^3+1/x+3/4x^(4/3)+c`

Explanation:

integrate each term using the `color(blue)"power rule for integration"`

`• int(ax^n)=a/(n+1)x^(n+1) ; n!=-1`

`rArrint(x^2-1/x^2+root(3)(x))dx`

`=int(x^2-x^-2+x^(1/3))larr" in exponent form"`

`=1/3x^3+1/x+3/4x^(4/3)+c`

where c is the constant of integration.