Search any question & find its solution
Question:
Answered & Verified by Expert
Integrate the function
$\frac{x e^x}{(1+x)^2}$
$\frac{x e^x}{(1+x)^2}$
Solution:
2861 Upvotes
Verified Answer
$\int \frac{(x+1-1) e^x}{(1+x)^2} d x=\int e^x\left(\frac{1}{x+1}-\frac{1}{(x+1)^2}\right) d x$
$=I_1-\int \frac{e^x}{(x+1)^2} d x \quad \ldots(i)$
$I_1=\frac{1}{1+x} \int e^x d x-\int\left(\frac{d}{d x}\left(\frac{1}{1+x}\right) \int e^x d x\right) d x$
$=\frac{e^x}{1+x}+\int \frac{e^x}{(1+x)^2} d x, \therefore I=I_1-\int \frac{e^x}{(1+x)^2} d x$
$=\frac{e^x}{1+x}+\int \frac{e^x}{(1+x)^2} d x-\int \frac{e^x}{(1+x)^2} d x=\frac{e^x}{1+x}+c$
$=I_1-\int \frac{e^x}{(x+1)^2} d x \quad \ldots(i)$
$I_1=\frac{1}{1+x} \int e^x d x-\int\left(\frac{d}{d x}\left(\frac{1}{1+x}\right) \int e^x d x\right) d x$
$=\frac{e^x}{1+x}+\int \frac{e^x}{(1+x)^2} d x, \therefore I=I_1-\int \frac{e^x}{(1+x)^2} d x$
$=\frac{e^x}{1+x}+\int \frac{e^x}{(1+x)^2} d x-\int \frac{e^x}{(1+x)^2} d x=\frac{e^x}{1+x}+c$
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.