Question

If $\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=?$

• A. $\frac{x^2}{\sqrt{x^2-1}}+c$
• B. $-\frac{x^2}{\sqrt{x^2-1}}+c$
• C. $\frac{\sqrt{x^2-1}}{x^2}+c$
• D. $-\frac{\sqrt{x^2-1}}{x^2}+c$

Answer 1

The correct option is D

$-\frac{\sqrt{x^2-1}}{x^2}+c$

Explanation for the correct option:

Step-1: Integrate the given expression

The given expression, $\frac{x^2-2}{x^3\sqrt{x^2-1}}=\frac{x^2}{x^3\sqrt{x^2-1}}-\frac{2}{x^3\sqrt{x^2-1}}$.

Integrate both sides of the equation.

$$\begin{gathered} \int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=\int \frac{x^{2}dx}{x^3\sqrt{x^2-1}}-\int \frac{2dx}{x^3\sqrt{x^2-1}} \\ =\int \frac{dx}{x\sqrt{x^2-1}}-2\int \frac{dx}{x^3\sqrt{x^2-1}} \end{gathered}$$

Let us assume that, $x=\sec \left(t\right)$.

Differentiate both sides of the equation with respect to $x$.

$dx=\sec \left(t\right)·\tan \left(t\right)dt$.

Thus, the integration of the given expression can be expressed as,

$\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=\int \frac{dx}{x\sqrt{x^2-1}}-2\int \frac{dx}{x^3\sqrt{x^2-1}}$

$$\begin{gathered} =\int \frac{\sec \left(t\right)·\tan \left(t\right)dt}{\sec \left(t\right)\sqrt{{\sec }^2\left(t\right)-1}}-2\int \frac{\sec \left(t\right)·\tan \left(t\right)dt}{{\sec }^3\left(t\right)\sqrt{{\sec }^2\left(t\right)-1}} \\ =\int \frac{\sec \left(t\right)·\tan \left(t\right)dt}{\sec \left(t\right)\sqrt{{\tan }^2\left(t\right)}}-2\int \frac{\sec \left(t\right)·\tan \left(t\right)dt}{{\sec }^3\left(t\right)\sqrt{{\tan }^2\left(t\right)}}\left[\because {\sec }^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1\right] \\ =\int \frac{\tan \left(t\right)dt}{\tan \left(t\right)}-2\int \frac{\tan \left(t\right)dt}{{\sec }^2\left(t\right)·\tan \left(t\right)} \\ =\int dt-2\int \frac{dt}{{\sec }^2\left(t\right)} \\ =t-\int 2{\cos }^2\left(t\right)dt\left[\because {\cos }^2\left(\theta \right)=\frac{1}{{\sec }^2\left(\theta \right)}\right] \\ =t-\int \left[\cos \left(2t\right)+1\right]dt\left[\because \cos \left(2\theta \right)+1=2{\cos }^2\left(t\right)\right] \\ =t-\int \cos \left(2t\right)dt-\int dt \\ =t-\frac{\sin \left(2t\right)}{2}-t+c \\ =-\frac{\sin \left(2t\right)}{2}+c \end{gathered}$$

Step 2: Simplify the derived expression

Now $x=\sec \left(t\right)$.

$$\begin{gathered} \Rightarrow \frac{1}{\sec \left(t\right)}=\frac{1}{x} \\ \Rightarrow \cos \left(t\right)=\frac{1}{x}\left[\because \cos \left(\theta \right)=\frac{1}{\sec \left(\theta \right)}\right] \\ \Rightarrow {\cos }^2\left(t\right)=\frac{1}{x^2} \\ \Rightarrow 1-{\sin }^2\left(t\right)=\frac{1}{x^2}\left[\because {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ \Rightarrow \sin \left(t\right)=\sqrt{1-\frac{1}{x^2}} \end{gathered}$$

Thus, $\sin \left(2t\right)=2\sin \left(t\right)·\cos \left(t\right)$

$$\begin{gathered} \Rightarrow \sin \left(2t\right)=2\times \sqrt{1-\frac{1}{x^2}}\times \frac{1}{x} \\ \Rightarrow \sin \left(2t\right)=2\times \sqrt{\frac{x^2-1}{x^2}}\times \frac{1}{x} \\ \Rightarrow \sin \left(2t\right)=\frac{2\sqrt{x^2-1}}{x}\times \frac{1}{x} \\ \Rightarrow \frac{\sin \left(2t\right)}{2}=\frac{\sqrt{x^2-1}}{x^2} \end{gathered}$$

Thus, $\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=-\frac{\sin \left(2t\right)}{2}+c$

$\Rightarrow \int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=-\frac{\sqrt{x^2-1}}{x^2}+c$

Hence, option(D) i.e. $-\frac{\sqrt{x^2-1}}{x^2}+c$ is the correct option.