First, we need to find the indefinite integral. ∫ _0^1xe^x(1+x)^2dx Apply Integration By Parts: u=xe^x, v'=1(1+x)^2 =xe^x( 11+x) ∫(e^x+e^xx)( 11+x ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

∫01xex1 + x2d...

Question

$\int_{0}^{1} \frac{x e^{x}}{(1 + x)^{2}} d x$ .

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Solution

First, we need to find the indefinite integral.
$\int_{0}^{1} \frac{x e^{x}}{{(1 + x)}^{2}} d x$
$A p p l y I n t e g r a t i o n B y P a r t s : u = x e^{x}, v^{'} = \frac{1}{{(1 + x)}^{2}}$
$= x e^{x} (- \frac{1}{1 + x}) - \int (e^{x} + e^{x} x) (- \frac{1}{1 + x}) d x$
$= - \frac{e^{x} x}{x + 1} - \int - \frac{e^{x} x + e^{x}}{x + 1} d x$
Now, we will evalaute the $\int - \frac{e^{x} x + e^{x}}{x + 1} d x$
$\int - \frac{e^{x} x + e^{x}}{x + 1} d x = - e^{x}$
So,
$= - \frac{e^{x} x}{x + 1} - (- e^{x})$

$= - \frac{e^{x} x}{x + 1} + e^{x} + C$
Now, we will evalaute the defnite integral

$\int_{0}^{1} \frac{x e^{x}}{{(1 + x)}^{2}} d x = - \frac{e}{2} + e - 1$

$= \frac{1}{2} (e - 2)$

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Q. Evaluate the definite integral

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$\int_{0}^{1} (x e^{x} + s i n \frac{π . x}{4}) d x$

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∫10xex(1+x)2dx.

$\int_{0}^{1} \frac{x e^{x}}{(1 + x)^{2}} d x$ .