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It is of integration please answer me as early as possible

It is of integration please answer me as early as possible

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Grade:12

2 Answers

Susmita
425 Points
5 years ago
Take a look at the part
\frac{1+x^2}{(1+x)^2}
=\frac{(1+x)^2 -2x}{(1+x)^2}
=1-\frac{2x}{(1+x)2}
=1-\frac{2x+2-2}{(1+x)2}
=1-\frac{2(x+1)}{(1+x)2} +\frac{2}{(1+x)^2}
=1-\frac{2}{(1+x)} +\frac{2}{(1+x)^2}
So the integration becomes
=\int e^x [1-\frac{2}{(1+x)} +\frac{2}{(1+x)^2}]dx
=\int e^x dx -2\int \frac{e^x dx}{1+x} +2\int \frac{e^x dx}{(1+x)^2}
The 1st term will give ex.We shall integrade the 2nd term in by parts taking ex as 2nd function.We shall keep the 3rd term untouched.
The 2nd term will yeild
-2[\frac{e^x}{1+x} + \int \frac{e^x dx}{(1+x)^2}]
The 2nd term of this part will cancel out with 3rd part of original integration.
So the answer will be
ex-2ex/(1+x)
Please approve the answer if you are helped
Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
562-228_integration by parts.JPG

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