Question

$\int \frac{1}{\sqrt{e^{2x}+4e^x+1}}dx$ is equal to:

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

Answer 1

The correct option is A $\ln |2+e^x+\sqrt{e^{2x}+4e^x+1}|+c$

We know that $\int \frac{dx}{\sqrt{x^2-a^2}}=\ln (x+\sqrt{x^2-a^2})+c$ ...... $\left(i\right)$

Let $I=\int \frac{1}{\sqrt{e^{2x}+4e^x+1}}dx$

$=\int \frac{1}{\sqrt{e^{2x}+4e^x+4-3}}dx$

$=\int \frac{1}{\sqrt{(e^x+2{)}^2-(\sqrt{3}{)}^2}}dx$

$=\ln |2+e^x+\sqrt{e^{2x}+4e^x+1}|+c$ ..... Using $\left(i\right)$