# How do you find the definite integral for: ((2x)/(1+x^2)^(1/3))dx for the intervals [0, sqrt7]?

Jan 1, 2017

First, find the indefinite integral and then evaluate from the lower bound to the upper bound.

#### Explanation:

int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = ?

Let $u = 1 + {x}^{2}$, then $\mathrm{du} = 2 x \mathrm{dx}$ and:

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

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

Reverse the substitution:

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

We do not need the constant for the definite integral:

${\int}_{0}^{\sqrt{7}} \frac{2 x}{1 + {x}^{2}} ^ \left(\frac{1}{3}\right) \mathrm{dx} = \frac{3}{2} {\left(1 + {x}^{2}\right)}^{\frac{2}{3}} {|}_{0}^{\sqrt{7}}$

${\int}_{0}^{\sqrt{7}} \frac{2 x}{1 + {x}^{2}} ^ \left(\frac{1}{3}\right) \mathrm{dx} = \frac{3}{2} {\left(1 + {\left(\sqrt{7}\right)}^{2}\right)}^{\frac{2}{3}} - \frac{3}{2} {\left(1 + {0}^{2}\right)}^{\frac{2}{3}}$

${\int}_{0}^{\sqrt{7}} \frac{2 x}{1 + {x}^{2}} ^ \left(\frac{1}{3}\right) \mathrm{dx} = \frac{3}{2} {\left(8\right)}^{\frac{2}{3}} - \frac{3}{2}$

${\int}_{0}^{\sqrt{7}} \frac{2 x}{1 + {x}^{2}} ^ \left(\frac{1}{3}\right) \mathrm{dx} = 4.5$