Q111 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$\int \frac{e^{2 x}}{\sqrt[4]{e^x+1}} d x=$
- A$\frac{4}{7}\left(e^x+1\right)^{4 / 3}\left(3 e^x-1\right)+c$
- B$\frac{2}{21}\left(e^x+1\right)^{3 / 4}\left(3 e^x-7\right)+c$
- Correct answerC$\frac{4}{21}\left(e^x+1\right)^{3 / 4}\left(3 e^x-4\right)+c$
- D$\frac{8}{21}\left(e^x+1\right)^{3 / 4}\left(3 e^x-1\right)+c$
Solution
$I=\int \frac{e^{2 x}}{\sqrt[4]{e^x+1}} d x$
Let $\left(e^x+1\right)^{\frac{1}{4}}=t \Rightarrow e^{2 x}=\left(t^4-1\right)^2$ Differentiating w.r.t. $x, 2 e^{2 \mathrm{x}} d x=2\left(t^4-1\right) 4 t^3 d t$
$\begin{aligned}
& \therefore \mathrm{I}=\int \frac{\left(t^4-1\right)\left(4 t^3\right) d t}{t} \Rightarrow \mathrm{I}=4 \int\left(t^4-1\right)\left(t^2\right) d t \\
& \Rightarrow \mathrm{I}=4 \int\left(t^6-t^2\right) d t \Rightarrow \mathrm{I}=4\left[\frac{t^7}{7}-\frac{t^3}{3}\right]+\mathrm{C}
\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{I}=4 t^3\left[\frac{t^4}{7}-\frac{1}{3}\right]+\mathrm{C} \\ & \Rightarrow \mathrm{I}=4\left(e^x+1\right)^{\frac{3}{4}}\left[\frac{\left(e^x+1\right)}{7}-\frac{1}{3}\right]+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\frac{4}{21}\left(e^x+1\right)^{\frac{3}{4}}\left(3 e^x-4\right)+\mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'C'.