Question

$\int \left\{\sqrt{\frac{\left(e^x+a\right)}{\left(e^x-a\right)}}\right\}dx=$

• A. $\cos h^{-1}\left(\frac{e^x}{a}\right)+se{c}^{-1}\left(\frac{e^x}{a}\right)+c$
• B. $\cos h^{-1}\left(\frac{e^x}{a}\right)-se{c}^{-1}\left(\frac{e^x}{a}\right)+c$
• C. $\frac{1}{a}\left[\cos h^{-1}\left(\frac{e^x}{a}\right)+se{c}^{-1}\left(\frac{e^x}{a}\right)\right]+c$
• D. $\frac{1}{a}\left[\cos h^{-1}\left(\frac{e^x}{a}\right)-se{c}^{-1}\left(\frac{e^x}{a}\right)\right]+c$

Answer 1

The correct option is A

$\cos h^{-1}\left(\frac{e^x}{a}\right)+se{c}^{-1}\left(\frac{e^x}{a}\right)+c$

Explanation for the correct option:

Finding the value of given integration:

Consider the given integral and multiply the denominator and numerator with $\sqrt{e^x+a}$,

$$\begin{gathered} I=\int \left\{\sqrt{\frac{\left(e^x+a\right)}{\left(e^x-a\right)}}\times \sqrt{\frac{e^x+a}{e^x+a}}\right\}dx \\ I=\int \frac{e^x}{\sqrt{e^{2x}-a^2}}dx+\int \frac{a{e}^x}{e^x\sqrt{e^{2x}-a^2}}dx \\ I=\int \frac{e^x}{\sqrt{e^{2x}-a^2}}dx+a\int \frac{e^x}{e^x\sqrt{e^{2x}-a^2}}dx \end{gathered}$$

Let us assume that

$$\begin{gathered} e^x=t \\ {e}^{x}dx=dt \end{gathered}$$[differentiating]

Then,

$I=\int \frac{1}{\sqrt{t^2-a^2}}dt+a\int \frac{1}{t\sqrt{t^2-a^2}}dt$

We know that,

$$\begin{gathered} \frac{d}{dx}\left(\cos h^{-1}\frac{x}{a}\right)=\frac{1}{\sqrt{x^2-a^2}} \\ \frac{d}{dx}\left(se{c}^{-1}\frac{x}{a}\right)=\frac{1}{x\sqrt{x^2-a^2}} \end{gathered}$$

So our integral value becomes,

$I=\cos h^{-1}\left(\frac{t}{a}\right)+se{c}^{-1}\left(\frac{t}{a}\right)+c$

Substitute the value of $t$ in the above equation becomes

$I=\cos h^{-1}\left(\frac{e^x}{a}\right)+se{c}^{-1}\left(\frac{e^x}{a}\right)+c$

Hence, the correct answer is Option (A).