Question

$\int e^x\sqrt{e^{2x}+1}dx$ is equal to
(where $C$ is integration constant)

• A. $\frac{e^x}{2}\sqrt{e^{2x}+1}+\frac{1}{2}\ln |e^x+\sqrt{e^{2x}+1}|+C$
• B. $\frac{1}{2}\ln |e^x+\sqrt{e^{2x}+1}|+C$
• C. $\frac{e^x}{2}\sqrt{e^x+1}+\frac{1}{2}\ln |e^x+\sqrt{e^{2x}+1}|+C$
• D. $\frac{1}{4}\ln \left|\frac{e^x+\sqrt{e^x+1}}{e^x-\sqrt{e^x+1}}\right|+C$

Answer 1

The correct option is A $\frac{e^x}{2}\sqrt{e^{2x}+1}+\frac{1}{2}\ln |e^x+\sqrt{e^{2x}+1}|+C$

Let $I=\int e^x\sqrt{e^{2x}+1}dx$
Let $e^x=t$
$\Rightarrow e^{x}dx=dt\Rightarrow I=\int \sqrt{t^2+1^2}dt=\frac{t}{2}\sqrt{t^2+1}+\frac{1}{2}\log |t+\sqrt{t^2+1}|+C\therefore I=\frac{e^x}{2}\sqrt{e^{2x}+1}+\frac{1}{2}\ln |e^x+\sqrt{e^{2x}+1}|+C[\because \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C]$