- 0 1 e x -x x-1 2 dx-

The correct option is C e2 1 = ∫_0^1 e^x[ 1x + 1 1( x + 1)^2]dx = [ e^xx + 1]^1_0 = e2 11 = e2 1 ...

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Standard XII

Mathematics

Integration Using Substitution

∫ 0 1 e x .x ...

Question

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

\frac{e}{2}

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1 + \frac{e}{2}

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\frac{e}{2} - 1

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1 - \frac{e}{2}

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Solution

The correct option is C $\frac{e}{2} - 1$
$= \int_{0}^{1} e^{x} [\frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}}] d x$
$= {[\frac{e^{x}}{x + 1}]}^{1}_{0}$
$= \frac{e}{2} - \frac{1}{1}$
$= \frac{e}{2} - 1$

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∫10ex.x(x+1)2dx=

The correct option is C e2−1=∫10ex[1x+1−1(x+1)2]dx=[exx+1]10=e2−11=e2−1

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

The correct option is C $\frac{e}{2} - 1$
$= \int_{0}^{1} e^{x} [\frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}}] d x$
$= {[\frac{e^{x}}{x + 1}]}^{1}_{0}$
$= \frac{e}{2} - \frac{1}{1}$
$= \frac{e}{2} - 1$