Question

Evaluate $\int \left[\sqrt{\frac{\left(x-1\right)}{\left(x+1\right)}}\right]dx$

• A. $2\sqrt{\left(x^2+1\right)}+{\sin }^{-1}x+c$
• B. $\sqrt{\left(x^2-1\right)}+{\sin }^{-1}x+c$
• C. $2\sqrt{\left(x^2-1\right)}+{\sin }^{-1}x+c$
• D. $2\sqrt{\left(x^2+1\right)}+\log \left|x+\sqrt{\left(x^2-1\right)}\right|+c$

Answer 1

The correct option is D

$2\sqrt{\left(x^2+1\right)}+\log \left|x+\sqrt{\left(x^2-1\right)}\right|+c$

Explanation for the correct option:

Evaluate the integral.

Step-1: Making numerator square root free:

$$\begin{gathered} \mathrm{I}=\int \left[\sqrt{\frac{\left(\mathrm{x}-1\right)}{\left(\mathrm{x}+1\right)}}\right]d\mathrm{x} \\ \mathrm{I}=\int \left[\frac{\sqrt{\left(\mathrm{x}-1\right)}}{\sqrt{\left(\mathrm{x}+1\right)}}\frac{\sqrt{\left(\mathrm{x}-1\right)}}{\sqrt{\left(\mathrm{x}-1\right)}}\right]d\mathrm{x}\left[\because \mathrm{Multiply}\mathrm{and}\mathrm{Divide}\mathrm{by}\sqrt{\left(\mathrm{x}-1\right)}\right] \\ \mathrm{I}=\int \frac{\left(\mathrm{x}-1\right)}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x} \\ \mathrm{I}=\int \left[\frac{\mathrm{x}}{\sqrt{\left({\mathrm{x}}^2+1\right)}}-\frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}\right]d\mathrm{x} \\ \mathrm{I}=\int \frac{\mathrm{x}}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}-\int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x} \\ \mathrm{I}=\frac{1}{2}\int \frac{2\mathrm{x}}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}-\int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}......\left(1\right)\left[\mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{by}\frac{1}{2}\right] \end{gathered}$$

Step 2: Using substitution method for integration:

Now consider,

$\int \frac{1}{\sqrt{\left(x^2+1\right)}}dx$

Let, $\mathrm{x}=\mathrm{sec\theta }$

Therefore, on differentiating we have,

$d\mathrm{x}=\mathrm{sec\theta }\mathrm{tan\theta }\mathrm{d\theta }$

Substituting the value, we get:

$$\begin{gathered} \int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}=\int \frac{\mathrm{sec\theta }\mathrm{tan\theta }d\mathrm{\theta }}{\sqrt{{\sec }^2\mathrm{\theta }-1}} \\ \int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}=\int \frac{\mathrm{sec\theta }\mathrm{tan\theta }d\mathrm{\theta }}{\sqrt{{\tan }^2\mathrm{\theta }}} \\ \int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}=\int \frac{\mathrm{sec\theta }\mathrm{tan\theta }d\mathrm{\theta }}{\mathrm{tan\theta }} \\ \int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}=\int \mathrm{sec\theta }d\mathrm{\theta } \\ \int \frac{1}{\sqrt{\left({\mathrm{x}}^2+1\right)}}d\mathrm{x}=\log \left|\mathrm{sec\theta }+\mathrm{tan\theta }\right|+\mathrm{c} \end{gathered}$$

Replacing $\theta$ term with $x$ term:

$$\begin{gathered} \Rightarrow \mathrm{x}=\mathrm{sec\theta } \\ \Rightarrow {\mathrm{x}}^2={\sec }^2\mathrm{\theta } \\ \Rightarrow {\mathrm{x}}^2={\tan }^2\mathrm{\theta }+1 \\ \Rightarrow {\mathrm{x}}^2-1={\tan }^2\mathrm{\theta } \\ \Rightarrow \sqrt{{\mathrm{x}}^2-1}=\mathrm{tan\theta } \end{gathered}$$

Then,

$\int \frac{1}{\sqrt{\left(x^2+1\right)}}dx=\log \left|x+\sqrt{x^2-1}\right|+c......\left(2\right)$

Step-3 : Using algebric substitution method for integration

Now considering $\int \frac{x}{\sqrt{\left(x^2+1\right)}}dx$

Let us assume that

$$\begin{gathered} t=x^2+1 \\ dt=2xdx \\ \frac{dt}{2x}=dx \end{gathered}$$

Then,

$$\begin{gathered} \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=\int \frac{dt}{\sqrt{t}} \\ \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=\int t^{-\frac{1}{2}}dt \end{gathered}$$

$$\begin{gathered} \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=\left(\frac{t^{-\frac{1}{2}+1}}{-{\displaystyle \frac{1}{2}}+1}\right)+c\left[\because \int x^{n}dx=\frac{x^{n+1}}{n+1}+c\right] \\ \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=\left(\frac{t^{\frac{1}{2}}}{{\displaystyle \frac{1}{2}}}\right)+c \\ \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=2\left(t^{\frac{1}{2}}\right)+c \\ \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=2{\left(x^2+1\right)}^{\frac{1}{2}}+c \\ \int \frac{2x}{\sqrt{\left(x^2+1\right)}}dx=2\left(\sqrt{x^2+1}\right)+c......\left(3\right) \end{gathered}$$

Substitute the value of the equation $\left(2\right)and\left(3\right)$ into the equation $\left(1\right)$ we get,

$I=2\sqrt{\left(x^2+1\right)}+\log \left|x+\sqrt{\left(x^2-1\right)}\right|+c$

Hence, option (D) is the correct answer.