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

Apr 13, 2017

$= {x}^{3} / 3 + \frac{1}{x} + \frac{3}{4} {x}^{\frac{4}{3}} + C$

#### Explanation:

$\int \left({x}^{2} - \frac{1}{x} ^ 2 + \sqrt[3]{x}\right) \mathrm{dx}$

By rewriting,

$= \int \left({x}^{2} - {x}^{- 2} + {x}^{\frac{1}{3}}\right) \mathrm{dx}$

By the Power Rule for integration, which states that $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$:

$= {x}^{3} / 3 - {x}^{- 1} / \left(- 1\right) + {x}^{\frac{4}{3}} / \left(\frac{4}{3}\right) + C$

By cleaning up,

$= {x}^{3} / 3 + \frac{1}{x} + \frac{3}{4} {x}^{\frac{4}{3}} + C$

I hope that this was clear.

Apr 13, 2017

$\frac{1}{3} {x}^{3} + \frac{1}{x} + \frac{3}{4} {x}^{\frac{4}{3}} + c$

#### Explanation:

integrate each term using the $\textcolor{b l u e}{\text{power rule for integration}}$

• int(ax^n)=a/(n+1)x^(n+1) ; n!=-1

$\Rightarrow \int \left({x}^{2} - \frac{1}{x} ^ 2 + \sqrt[3]{x}\right) \mathrm{dx}$

$= \int \left({x}^{2} - {x}^{-} 2 + {x}^{\frac{1}{3}}\right) \leftarrow \text{ in exponent form}$

$= \frac{1}{3} {x}^{3} + \frac{1}{x} + \frac{3}{4} {x}^{\frac{4}{3}} + c$

where c is the constant of integration.