Question

$I=\int \frac{e^x}{(e^x+1{)}^{\frac{1}{4}}}dx$ is equal to

• A. $\frac{e^{2x}-e^{-2x}}{2}+c$
• B. `$\frac{e^{2x}+e^{-2x}}{2}+c$
• C. $\frac{e^x-e^{2x}}{2}+c$
• D. none of these

Answer 1

The correct option is D none of these

$I=\int \frac{e^x}{(e^x+1{)}^{\frac{1}{4}}}dx$

Let $e^x+1=t$

$e^{x}dx=dt$

$\therefore I=\int \frac{dt}{t^{\frac{1}{4}}}=\int t^{\frac{-1}{4}}dt$

$=\frac{t^{\frac{3}{4}}}{\frac{3}{4}}=\frac{4}{3}\left(t^{\frac{3}{4}}\right)$

$I=\frac{4}{3}(e^x+1{)}^{\frac{3}{4}}$