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

Explanation:

Step1: Use substitution

Let $u = 2x$, then $du=2dx$ and $x=\frac{u}{2}$. The integral becomes $\int\frac{du/2}{\frac{u}{2}\sqrt{9 + u^{2}}}=\int\frac{du}{u\sqrt{9 + u^{2}}}$.

Step2: Use another substitution

Let $u = 3\tan\theta$, then $du=3\sec^{2}\theta d\theta$. And $\sqrt{9 + u^{2}}=\sqrt{9+9\tan^{2}\theta}=3\sec\theta$. The integral is $\int\frac{3\sec^{2}\theta d\theta}{3\tan\theta\cdot3\sec\theta}=\frac{1}{3}\int\frac{\sec\theta}{\tan\theta}d\theta$.

Step3: Simplify the integrand

Since $\frac{\sec\theta}{\tan\theta}=\frac{1}{\sin\theta}$, the integral is $\frac{1}{3}\int\frac{d\theta}{\sin\theta}=\frac{1}{3}\int\csc\theta d\theta$.

Step4: Recall the integral of $\csc\theta$

The integral of $\csc\theta$ is $\ln|\csc\theta-\cot\theta|+C$.

Step5: Back - substitute

Since $u = 3\tan\theta$, $\tan\theta=\frac{u}{3}$, $\csc\theta=\frac{\sqrt{u^{2}+9}}{u}$ and $\cot\theta=\frac{3}{u}$. The result is $\frac{1}{3}\ln\left|\frac{\sqrt{u^{2}+9}}{u}-\frac{3}{u}\right|+C$.

Step6: Back - substitute $u = 2x$

We get $\frac{1}{3}\ln\left|\frac{\sqrt{4x^{2}+9}}{2x}-\frac{3}{2x}\right|+C=\frac{1}{3}\ln\left|\frac{\sqrt{4x^{2}+9}-3}{2x}\right|+C$.

Answer:

$\frac{1}{3}\ln\left|\frac{\sqrt{4x^{2}+9}-3}{2x}\right|+C$