Question

$\int \sqrt{\frac{e^x-1}{e^x+1}}dx=$

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

Answer 1

The correct option is A $\log (e^x+\sqrt{e^{2x}-1})-{\sec }^{-1}\left(e^x\right)+c$

$\int \frac{e^x-1}{\sqrt{e^{2x}-1}}dx.$

$\int \frac{e^x}{\sqrt{e^{2x}-1}}\cdot dx-\int \frac{1}{\sqrt{e^{2x}-1}}\cdot dx$

$\frac{d{e}^x}{\sqrt{e^{2x}-1}}-\int \frac{1}{\sqrt{e^{2x}-1}}dx.\int \frac{1}{\sqrt{e^{2x}-1}}dx.$

$log\mid e^x+\sqrt{e^{2x}-1}\mid -\int \frac{1}{\sqrt{e^{2x}-1}}dx$

Let $e^x=t.$

$e^{x}dx=dt$

$dx=\frac{1}{t}dt$

$log\mid e^x+\sqrt{e^{2x}-1}\mid -\int \frac{dt}{t\sqrt{t^2-1}}$

$log\mid e^x+\sqrt{e^{2x}-1}\mid -se{c}^{-1}\left(t\right)+c$

$=log\mid e^x+\sqrt{e^{2x}-1}\mid -se{c}^{-1}\left(e^x\right)-c.$