Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{x^{2 x}}{(1+2 x)} d x=$
(where C is a constant of integration.)
Options:
(where C is a constant of integration.)
Solution:
2154 Upvotes
Verified Answer
The correct answer is:
$\frac{e^{2 x}}{4(1+2 x)}+C$
$\int \frac{x \cdot e^{2 x}}{(1+2 x)^2} d x$
Let $2 \mathrm{x}=\mathrm{t} \Rightarrow 2 \mathrm{dx}=\mathrm{dt}$
$\begin{aligned} & =\frac{1}{4} \int \frac{2 \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}}}{(1+2 \mathrm{x})^2} \cdot 2 \mathrm{dx} \\ & =\frac{1}{4} \int \frac{\mathrm{te} \mathrm{dt}}{(1+\mathrm{t})^2} \\ & =\frac{1}{4} \int \mathrm{e}^{\mathrm{t}}\left\{\frac{1}{1+\mathrm{t}}-\frac{1}{(1+\mathrm{t})^2}\right\} \mathrm{dt} \\ & =\frac{1}{4} \mathrm{e}^{\mathrm{t}} \cdot \frac{1}{1+\mathrm{t}}+\mathrm{C}\left[\because \int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{dx}=\mathrm{e}^{\mathrm{xf}}(\mathrm{x})+\mathrm{C}\right]\end{aligned}$
Let $2 \mathrm{x}=\mathrm{t} \Rightarrow 2 \mathrm{dx}=\mathrm{dt}$
$\begin{aligned} & =\frac{1}{4} \int \frac{2 \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}}}{(1+2 \mathrm{x})^2} \cdot 2 \mathrm{dx} \\ & =\frac{1}{4} \int \frac{\mathrm{te} \mathrm{dt}}{(1+\mathrm{t})^2} \\ & =\frac{1}{4} \int \mathrm{e}^{\mathrm{t}}\left\{\frac{1}{1+\mathrm{t}}-\frac{1}{(1+\mathrm{t})^2}\right\} \mathrm{dt} \\ & =\frac{1}{4} \mathrm{e}^{\mathrm{t}} \cdot \frac{1}{1+\mathrm{t}}+\mathrm{C}\left[\because \int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{dx}=\mathrm{e}^{\mathrm{xf}}(\mathrm{x})+\mathrm{C}\right]\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.