$int\frac{x^{2}dx}{(x^{2}-a^{2})^{\frac{3}{2}}}$

Explanation:

Step1: Use trigonometric substitution

Let $x = a\sec\theta$, then $dx=a\sec\theta\tan\theta d\theta$. And $x^{2}-a^{2}=a^{2}\sec^{2}\theta - a^{2}=a^{2}\tan^{2}\theta$. $$\int\frac{x^{2}dx}{(x^{2}-a^{2})^{\frac{3}{2}}}=\int\frac{a^{2}\sec^{2}\theta\cdot a\sec\theta\tan\theta d\theta}{(a^{2}\tan^{2}\theta)^{\frac{3}{2}}}$$

Step2: Simplify the integrand

$$ \begin{align*} \int\frac{a^{2}\sec^{2}\theta\cdot a\sec\theta\tan\theta d\theta}{(a^{2}\tan^{2}\theta)^{\frac{3}{2}}}&=\int\frac{a^{3}\sec^{3}\theta\tan\theta d\theta}{a^{3}\tan^{3}\theta}\ &=\int\frac{\sec^{3}\theta}{\tan^{2}\theta}d\theta\ &=\int\frac{\frac{1}{\cos^{3}\theta}}{\frac{\sin^{2}\theta}{\cos^{2}\theta}}d\theta\ &=\int\frac{1}{\cos\theta\sin^{2}\theta}d\theta \end{align*} $$

Step3: Use another substitution

Let $u = \sin\theta$, then $du=\cos\theta d\theta$. And $\frac{1}{\cos\theta\sin^{2}\theta}=\frac{1}{(1 - u^{2})u^{2}}$. Decompose $\frac{1}{(1 - u^{2})u^{2}}=\frac{1}{(1 - u)(1 + u)u^{2}}$ into partial - fractions: $\frac{1}{(1 - u)(1 + u)u^{2}}=\frac{1}{2u^{2}}+\frac{1}{4(1 - u)}+\frac{1}{4(1 + u)}$. $$\int\frac{1}{\cos\theta\sin^{2}\theta}d\theta=\int\left(\frac{1}{2u^{2}}+\frac{1}{4(1 - u)}+\frac{1}{4(1 + u)}\right)du$$

Step4: Integrate term - by - term

$$ \begin{align*} \int\left(\frac{1}{2u^{2}}+\frac{1}{4(1 - u)}+\frac{1}{4(1 + u)}\right)du&=-\frac{1}{2u}-\frac{1}{4}\ln|1 - u|+\frac{1}{4}\ln|1 + u|+C\ &=-\frac{1}{2\sin\theta}-\frac{1}{4}\ln\left|\frac{1 - \sin\theta}{1+\sin\theta}\right|+C \end{align*} $$ Since $x = a\sec\theta$, then $\cos\theta=\frac{a}{x}$ and $\sin\theta=\frac{\sqrt{x^{2}-a^{2}}}{x}$.

Step5: Back - substitute

$$ \begin{align*} &-\frac{x}{2\sqrt{x^{2}-a^{2}}}-\frac{1}{4}\ln\left|\frac{x-\sqrt{x^{2}-a^{2}}}{x + \sqrt{x^{2}-a^{2}}}\right|+C\ &=-\frac{x}{2\sqrt{x^{2}-a^{2}}}+\frac{1}{2}\text{arcosh}\left(\frac{x}{a}\right)+C \end{align*} $$

Answer:

$-\frac{x}{2\sqrt{x^{2}-a^{2}}}+\frac{1}{2}\text{arcosh}\left(\frac{x}{a}\right)+C$