# How do you find the antiderivative of int (6x)/(x^2-7)^(1/9)dx from [3,4]?

Feb 1, 2017

The integral is approximately equal to $17.546$.

#### Explanation:

The first thing to do with definite integrals is to make sure that they're in fact definite and not improper.

The integral $\frac{6 x}{{x}^{2} - 7} ^ \left(\frac{1}{9}\right)$ is continuous on $\left[3 , 4\right]$. We are dealing with a definite integral.

I think we should consider a u-substitution to integrate. Let $u = {x}^{2} - 7$. Then $\mathrm{du} = 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{2 x}$. Furthermore, the new bounds of integration become $2$ to $9$ because we will now be working in $u$.

$\implies {\int}_{2}^{9} \frac{6 x}{u} ^ \left(\frac{1}{9}\right) \cdot \frac{\mathrm{du}}{2 x}$

$\implies {\int}_{2}^{9} \frac{3}{u} ^ \left(\frac{1}{9}\right)$

$\implies 3 {\int}_{2}^{9} {u}^{- \frac{1}{9}}$

$\implies 3 {\left[\frac{9}{8} {u}^{\frac{8}{9}}\right]}_{2}^{9}$

$\implies 3 \left[\frac{9}{8} {\left(9\right)}^{\frac{8}{9}} - \frac{9}{8} {\left(2\right)}^{\frac{8}{9}}\right]$

$\approx 17.546$

Hopefully this helps!