The correct option is B e/2 1 ∫_0^1e^x ( 1(x=+1) 1(x+1)^2 ) dx It is in the form of ∫ #160; e^x [ f(x)+f^1 (x) ] #160; dx=e^x f(x)+c ...

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Integration Using Substitution

∫01xex/x+12dx...

Question

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

\frac{e}{2}

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\frac{e - 1}{2}

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\frac{e}{2} - 1

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\frac{e - 3}{2}

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Solution

The correct option is B $\frac{e}{2} - 1$
$\int_{0}^{1} e^{x} (\frac{1}{(x = + 1)} - \frac{1}{(x + 1)^{2}}) d x$
It is in the form of
$\int e^{x} [f (x) + f^{1} (x)] d x = e^{x} f (x) + c$
$= {[\frac{e^{x}}{(1 + x)} + c]}_{0}^{1}$
$= \frac{e}{2} - 1$
$= \frac{e}{2} - 1$

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Similar questions

Q. The integral

e \int 1 {{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{x}} {log}_{e} x d x

is equal to:

$I f e_{1} a n d e_{2} a r e r e s p e c t i v e l y t h e e c c e n t r i c i t i e s o f t h e e l l i p s e \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1 a n d t h e h y p e r b o l a \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1, t h e n t h e r e l a t i o n b e t w e e n e_{1} a n d e_{2} i s$

$I f e_{1} i s t h e e c c e n t r i c i t y o f t h e c o n i c 9 x^{2} + 4 y^{2} = 36 a n d e^{2} i s t h e e c c e n t r i c i t y o f t h e c o n i c 9 x^{2} - 4 y^{2} = 36, t h e n$

$\int x^{2} e^{x^{3}} d x =$

The solution of the differential equation $log (\frac{d y}{d x}) = 4 x - 2 y - 2$ , y = 1 when x = 1 is:

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∫10xex(x+1)2dx=

The correct option is B e2−1 ∫10ex(1(x=+1)−1(x+1)2)dxIt is in the form of∫ex[f(x)+f1(x)]dx=exf(x)+c=[ex(1+x)+c]10=e2−1=e2−1

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

The correct option is B $\frac{e}{2} - 1$
$\int_{0}^{1} e^{x} (\frac{1}{(x = + 1)} - \frac{1}{(x + 1)^{2}}) d x$
It is in the form of
$\int e^{x} [f (x) + f^{1} (x)] d x = e^{x} f (x) + c$
$= {[\frac{e^{x}}{(1 + x)} + c]}_{0}^{1}$
$= \frac{e}{2} - 1$
$= \frac{e}{2} - 1$