# How can i solve this integral? int (2/(5x)sqrt (3+4/x^2)dx

Sep 2, 2017

I got: $\frac{2 \sqrt{3}}{5} \ln \left\mid \frac{3 {x}^{2} + 4}{2 \sqrt{3} x} + \frac{\sqrt{3} x}{2} - \frac{2}{\sqrt{3} x} \right\mid + C$

#### Explanation:

$I = \int \frac{2}{5 x} \sqrt{3 + \frac{4}{x} ^ 2} \mathrm{dx}$

Use the substitution $x = \frac{2}{\sqrt{3}} \cot \theta$. This implies that $\mathrm{dx} = \frac{- 2}{\sqrt{3}} \cot \theta \csc \theta d \theta$.

$I = \frac{2}{5} \int \frac{\sqrt{3}}{2} \tan \theta \sqrt{3 + 4 {\left(\frac{\sqrt{3}}{2} \tan \theta\right)}^{2}} \left(\frac{- 2}{\sqrt{3}} \cot \theta \csc \theta d \theta\right)$

Canceling $\frac{\sqrt{3}}{2} \tan \theta \left(\frac{2}{\sqrt{3}} \cot \theta\right) = 1$:

$I = \frac{- 2}{5} \int \csc \theta \sqrt{3 + 3 {\tan}^{2} \theta} d \theta$

Factoring $\sqrt{3}$, the radical becomes $\sqrt{1 + {\tan}^{2} \theta} = \sec \theta$:

$I = \frac{- 2 \sqrt{3}}{5} \int \frac{d \theta}{\sin \theta \cos \theta}$

Note that $\sin \theta \cos \theta = \frac{1}{2} \sin 2 \theta$:

$I = \frac{- 4 \sqrt{3}}{5} \int \frac{d \theta}{\sin 2 \theta}$

$I = \frac{- 4 \sqrt{3}}{5} \int \csc 2 \theta d \theta$

The integral of cosecant is fairly standard. Look it up here if you aren't sure.

Note that I'll use the form $\int \csc x = - \ln \left\mid \csc x + \cot x \right\mid$, which is equivalent to $\ln \left\mid \csc x - \cot x \right\mid$. I'm choosing the first one so the minus sign will cancel with the minus sign already outside the integral.

You'll also need to undo the $2 \theta$ by multiplying the integral by $1 / 2$.

$I = \frac{2 \sqrt{3}}{5} \ln \left\mid \csc 2 \theta + \cot 2 \theta \right\mid + C$

Use $\csc 2 \theta = \frac{1}{\sin 2 \theta} = \frac{1}{2 \sin \theta \cos \theta}$ and $\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$:

$I = \frac{2 \sqrt{3}}{5} \ln \left\mid \frac{1}{2 \sin \theta \cos \theta} + \frac{1 - {\tan}^{2} \theta}{2 \tan \theta} \right\mid + C$

Before proceeding we can factor $2$ from both these denominators and remove $\frac{- 2 \sqrt{3}}{5} \ln 2$ from the logarithm using $\log \left(A / B\right) = \log A - \log B$. This constant is absorbed into $C$.

$I = \frac{2 \sqrt{3}}{5} \ln \left\mid \csc \theta \sec \theta + \cot \theta - \tan \theta \right\mid + C$

Our original substitution was $\cot \theta = \frac{\sqrt{3} x}{2}$. Thus:

• $\tan \theta = \frac{1}{\cot} \theta = \frac{2}{\sqrt{3} x}$
• $\csc \theta = \sqrt{{\cot}^{2} \theta + 1} = \frac{\sqrt{3 {x}^{2} + 4}}{2}$
• $\sec \theta = \sqrt{{\tan}^{2} \theta + 1} = \frac{\sqrt{3 {x}^{2} + 4}}{\sqrt{3} x}$

So:

$I = \frac{2 \sqrt{3}}{5} \ln \left\mid \frac{3 {x}^{2} + 4}{2 \sqrt{3} x} + \frac{\sqrt{3} x}{2} - \frac{2}{\sqrt{3} x} \right\mid + C$