MCQ
$\int \limits_0^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^x+6} d x$
  • A
    $\log _e\left(\frac{512}{81}\right)$
  • $\log _e\left(\frac{32}{27}\right)$
  • C
    $\log _e\left(\frac{256}{81}\right)$
  • D
    $\log _e\left(\frac{64}{27}\right)$

Answer

Correct option: B.
$\log _e\left(\frac{32}{27}\right)$
b
$1=\int \limits_0^{\infty} \frac{6}{\left(e^x+1\right)\left(e^{ x }+2\right)\left( e ^{ x }+3\right)} dx$

$=6 \int \limits_0^{\infty}\left(\frac{\frac{1}{2}}{ e ^{ x }+1}+\frac{-1}{ e ^{ x }+2}+\frac{\frac{1}{2}}{ e ^{ x }+3}\right) d x$

$=3 \int \limits_0^{\infty} \frac{ e ^{- x }}{1+ e ^{- x }} dx -6 \int \limits_0^{\infty} \frac{ e ^{- x } dx }{1+2 e ^{- x }}+3 \int \limits_0^{\infty} \frac{ e ^{- x }}{1+3 e ^{- x }} dx$

$=3\left[-\ln \left(1+ e ^{- x }\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 e ^{- x }\right)\right]_0^{\infty}$

$-\frac{3}{3}\left[\ln \left(1+3 e ^{- x }\right)\right]_0^{\infty}$

$=3 \ln 2-3 \ln 3+\ln 4$

$=3 \ln \frac{2}{3}+\ln 4$

$=\ln \frac{32}{27}$

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