# Question #2a34f

Aug 31, 2017

$\int {x}^{2} / \left(9 + 16 {x}^{6}\right) \mathrm{dx} = \frac{1}{36} \arctan \left(\frac{4}{3} {x}^{3}\right) + \text{constant}$

#### Explanation:

For the integral $F \left(x\right) = \int {x}^{2} / \left(9 + 16 {x}^{6}\right) \mathrm{dx}$, sub $u = {x}^{3}$ and $\mathrm{du} = 3 {x}^{2} \mathrm{dx}$

Then $F \left(x\right) = \frac{1}{3} \int \frac{1}{9 + 16 {u}^{2}} \mathrm{du} = \frac{1}{27} \int \frac{1}{1 + \frac{16}{9} {u}^{2}} \mathrm{du}$

For the integral $F \left(x\right)$, sub $t = \frac{4}{3} u$ and $\mathrm{dt} = \frac{4}{3} \mathrm{du}$

Then $F \left(x\right) = \frac{1}{36} \int \frac{1}{1 + {t}^{2}} \mathrm{dt}$

$F \left(x\right)$ is now in a standard form and can be evaluated as a standard integral.

$F \left(x\right) = \frac{1}{36} \int \frac{1}{1 + {t}^{2}} \mathrm{dt} = \frac{1}{36} \arctan t = \frac{1}{36} \arctan \left(\frac{4}{3} u\right) = \frac{1}{36} \arctan \left(\frac{4}{3} {x}^{3}\right) + \text{constant}$

Sep 1, 2017

$\frac{1}{36} {\tan}^{-} 1 \left(\frac{4}{3} {x}^{3}\right) + C$

#### Explanation:

$\int \frac{{x}^{2} \mathrm{dx}}{9 + 16 {x}^{6}}$

Use the substitution ${x}^{3} = \frac{3}{4} \tan \theta , \implies 3 {x}^{2} \mathrm{dx} = \frac{3}{4} {\sec}^{2} \theta d \theta$.

$= \frac{1}{3} \int \frac{3 {x}^{2} \mathrm{dx}}{9 + 16 {\left({x}^{3}\right)}^{2}} = \frac{1}{3} \int \frac{\frac{3}{4} {\sec}^{2} \theta d \theta}{9 + 16 \left(\frac{9}{16} {\tan}^{2} \theta\right)}$

$= \frac{1}{4} \int \frac{{\sec}^{2} \theta d \theta}{9 \left(1 + {\tan}^{2} \theta\right)}$

Recall that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$= \frac{1}{36} \int d \theta = \frac{1}{36} \theta + C$

From ${x}^{3} = \frac{3}{4} \tan \theta$, note that $\tan \theta = \frac{4}{3} {x}^{3}$ so $\theta = {\tan}^{-} 1 \left(\frac{4}{3} {x}^{3}\right)$:

$= \frac{1}{36} {\tan}^{-} 1 \left(\frac{4}{3} {x}^{3}\right) + C$