# How do you integrate int dx/(4x^2+9)^2 using trig substitutions?

Mar 21, 2018

$\int \frac{\mathrm{dx}}{4 {x}^{2} + 9} ^ 2 = \frac{1}{108} \arctan \left(\frac{2 x}{3}\right) + \frac{x}{72 {x}^{2} + 162} + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{4 {x}^{2} + 9} ^ 2$

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

After using $2 x = 3 \tan u$ and $2 \mathrm{dx} = 3 {\left(\sec u\right)}^{2} \cdot \mathrm{du}$ transforms, this integral became

$\frac{1}{2} \int \frac{3 {\left(\sec u\right)}^{2} \cdot \mathrm{du}}{81 {\left(\sec u\right)}^{4}}$

=$\frac{1}{54} \int {\left(\cos u\right)}^{2} \cdot \mathrm{du}$

=$\frac{1}{108} \int \left(1 + \cos 2 u\right) \cdot \mathrm{du}$

=$\frac{1}{108} u + \frac{1}{216} \sin 2 u + C$

=$\frac{1}{108} u + \frac{1}{216} \cdot \frac{2 \tan u}{{\left(\tan u\right)}^{2} + 1} + C$

After using $2 x = 3 \tan u$, $\tan u = \frac{2 x}{3}$ and $u = \arctan \left(\frac{2 x}{3}\right)$ inverse transforms, I found

$\int \frac{\mathrm{dx}}{4 {x}^{2} + 9} ^ 2$

=$\frac{1}{108} \arctan \left(\frac{2 x}{3}\right) + \frac{1}{216} \cdot \frac{2 \cdot \frac{2 x}{3}}{{\left(\frac{2 x}{3}\right)}^{2} + 1} + C$

=$\frac{1}{108} \arctan \left(\frac{2 x}{3}\right) + \frac{x}{72 {x}^{2} + 162} + C$