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

Explanation:

Step1: Use substitution

Let $x = 2\tan\theta$, then $dx=2\sec^{2}\theta d\theta$. And $\sqrt{x^{2}+4}=\sqrt{4\tan^{2}\theta + 4}=2\sec\theta$.

Step2: Rewrite the integral

Substitute $x$ and $dx$ into the integral: [ \begin{align*} \int\frac{dx}{x\sqrt{x^{2}+4}}&=\int\frac{2\sec^{2}\theta d\theta}{2\tan\theta\times2\sec\theta}\ &=\frac{1}{2}\int\frac{\sec\theta}{\tan\theta}d\theta\ &=\frac{1}{2}\int\frac{1}{\sin\theta}d\theta\ &=\frac{1}{2}\int\csc\theta d\theta \end{align*} ]

Step3: Integrate $\csc\theta$

The integral of $\csc\theta$ is $\ln|\csc\theta-\cot\theta| + C$. So $\frac{1}{2}\int\csc\theta d\theta=\frac{1}{2}\ln|\csc\theta-\cot\theta|+C$.

Step4: Back - substitute

Since $x = 2\tan\theta$, then $\tan\theta=\frac{x}{2}$. We know that $\csc\theta=\frac{\sqrt{x^{2}+4}}{x}$ and $\cot\theta=\frac{2}{x}$. [ \begin{align*} \frac{1}{2}\ln|\csc\theta-\cot\theta|&=\frac{1}{2}\ln\left|\frac{\sqrt{x^{2}+4}}{x}-\frac{2}{x}\right|\ &=\frac{1}{2}\ln\left|\frac{\sqrt{x^{2}+4}-2}{x}\right|+C \end{align*} ]

Answer:

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