∫0∞6e3x+6e2x+11ex+6dx=

Put ex=t⇒dx=dtt x=0⇒t=1 x→∞⇒ t→∞ now I=∫1∞6t(t3+6t2+11t+6)dt=∫1∞6t(t+1) (t+2)(t+3)dt ∫1∞(1t−3t+1+3t+2−1t+3)dt=[lnt−3ln(t+1)+3ln(t+2)−ln(t+3)]1∞=[lnt(t+2)3(t+1)3 (t+3)]1∞=ltt→∞ ln (t(t+2)3(t+1)3 (t+3))−ln (1.3323.4)=0−ln(2732)=ln(3227)

∫0∞6e3x+6e2x+11ex+6dx=

Put ex=t⇒dx=dtt x=0⇒t=1 x→∞⇒ t→∞ now I=∫1∞6t(t3+6t2+11t+6)dt=∫1∞6t(t+1) (t+2)(t+3)dt ∫1∞(1t−3t+1+3t+2−1t+3)dt=[lnt−3ln(t+1)+3ln(t+2)−ln(t+3)]1∞=[lnt(t+2)3(t+1)3 (t+3)]1∞=ltt→∞ ln (t(t+2)3(t+1)3 (t+3))−ln (1.3323.4)=0−ln(2732)=ln(3227)

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$\int_{0}^{\infty} \frac{6}{e^{3 x} + 6 e^{2 x} + 11 e^{x} + 6} dx =$

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$\log_{e} (\frac{64}{27})$

$\log_{e} (\frac{512}{81})$

$\log_{e} (\frac{32}{27})$

$\log_{e} (\frac{256}{81})$

answer is C.

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Detailed Solution

Put
$e^{x} = t \Rightarrow dx = \frac{dt}{t} x = 0 \Rightarrow t = 1 x \to \infty \Rightarrow t \to \infty now I = \int_{1}^{\infty} \frac{6}{t (t^{3} + 6 t^{2} + 11 t + 6)} dt$
$= \int_{1}^{\infty} \frac{6}{t (t + 1) (t + 2) (t + 3)} dt \int_{1}^{\infty} (\frac{1}{t} - \frac{3}{t + 1} + \frac{3}{t + 2} - \frac{1}{t + 3}) dt$
$= {[lnt - 3 \ln (t + 1) + 3 \ln (t + 2) - \ln (t + 3)]}_{1}^{\infty} = {[\ln \frac{t {(t + 2)}^{3}}{{(t + 1)}^{3} (t + 3)}]}_{1}^{\infty}$
$= \underset{t \to \infty}{lt} \ln (\frac{t {(t + 2)}^{3}}{{(t + 1)}^{3} (t + 3)}) - \ln (\frac{1 . 3^{3}}{2^{3} . 4}) = 0 - \ln (\frac{27}{32}) = \ln (\frac{32}{27})$