What is $\int_{- \infty}^{\infty} \frac{e^{x}}{(e^{2 x}) + 3} d x$ $int_-oo^oo (e^x)/((e^(2x))+3) dx$ ?

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What is $\int_{- \infty}^{\infty} \frac{e^{x}}{(e^{2 x}) + 3} d x$ $int_-oo^oo (e^x)/((e^(2x))+3) dx$ ?

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Answer 1 · 2016-03-24T14:51:45

$\frac{π}{\sqrt{3}}$ $pi/sqrt3$

Explanation:

$\int_{- \infty}^{\infty} \frac{e^{x}}{e^{2 x} + 3} d x =$ $int_-oo^oo e^x/(e^(2x)+3)dx=$

Using trigonometric substitution
$e^{x} = \sqrt{3} tan y$ $e^x=sqrt3tany$
$e^{x} \cdot d x = \sqrt{3} {sec}^{2} y \cdot d y$ $e^x*dx=sqrt3sec^2y*dy$

Since $tan y = \frac{e^{x}}{\sqrt{3}}$ $tany=e^x/sqrt3$ , when
$x \to - \infty$ $x->-oo$ => $tan y \to - \infty$ $tany -> -oo$ => $y \to {(- \frac{π}{2})}^{+}$ $y->(-pi/2)^"+"$
$x \to \infty$ $x->oo$ => $tan y \to \infty$ $tany -> oo$ => $y \to {(\frac{π}{2})}^{-}$ $y->(pi/2)^"-"$

So the expression becomes

$=int_((-pi/2)^"+")^((pi/2)^"-") (sqrt3sec^2y)/(3tan^2y+3)dy=int_((-pi/2)^"+")^((pi/2)^"-") (sqrt3cancel(sec^2y))/(3cancel(sec^2y))dy$
$=(y/sqrt3)|_((-pi/2)^"+")^((pi/2)^"-")$
$=1/sqrt3[pi/2-(-pi/2)]=pi/sqrt3$

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What is $\int_{- \infty}^{\infty} \frac{e^{x}}{(e^{2 x}) + 3} d x$ $int_-oo^oo (e^x)/((e^(2x))+3) dx$ ?

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Explanation:

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