The correct option is C 1/4 ∫_0^1x #160; #10;dx(x^2+1)^2 #10; ∫x #160; dx(x^2+1)^2 =12∫ #10;dx^2(1+x^2)^2= 12 ( 11+x^2 )^ 1 #10; ...

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Integration Using Substitution

∫01xdx/x2+12=

Question

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

1 / 2

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1 / 4

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Solution

The correct option is C $1 / 4$
$\int_{0}^{1} \frac{x d x}{(x^{2} + 1)^{2}}$
$\int \frac{x d x}{(x^{2} + 1)^{2}} = \frac{1}{2} \int \frac{d x^{2}}{(1 + x^{2})^{2}} = - \frac{1}{2} {(\frac{1}{1 + x^{2}})}^{- 1}$
$= - \frac{1}{2} (1 + x^{2})^{+ c}$
$= \int_{0}^{1} \frac{x d x}{(x^{2} + 1)^{2}} = [- \frac{1}{2} (1 + x^{2}) + c] \int_{0}^{1}$
$= [- \frac{1}{2} (2) + c] - [- \frac{1}{2} + c]$
$= - \frac{1}{4} + \frac{1}{2} = \frac{1}{4}$
$\int_{0}^{1} \frac{x d x}{(x^{2} + 1)^{2}} = \frac{1}{4}$

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