int limits_0^{infty} frac{6}{e^{3 x}+6 e^{2 x | vedclass

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(B) Let $I = \int \limits_0^{\infty} \frac{6}{(e^x+1)(e^x+2)(e^x+3)} dx$. Using partial fractions,we write: $\frac{6}{(e^x+1)(e^x+2)(e^x+3)} = \frac{3

Vedclass · Accepted Answer

$\log _e\left(\frac{32}{27}\right)$

Vedclass · Answer

(B) Let $I = \int \limits_0^{\infty} \frac{6}{(e^x+1)(e^x+2)(e^x+3)} dx$. Using partial fractions,we write: $\frac{6}{(e^x+1)(e^x+2)(e^x+3)} = \frac{3}{e^x+1} - \frac{6}{e^x+2} + \frac{3}{e^x+3}$. Substituting $u = e^x$,then $du = e^x dx$,so $dx = \frac{du}{u}$. $I = \int_1^{\infty} \frac{6}{u(u+1)(u+2)(u+3)} du$. Using partial fractions for the integrand: $\frac{6}{u(u+1)(u+2)(u+3)} = \frac{1}{u} - \frac{3}{u+1} + \frac{3}{u+2} - \frac{1}{u+3}$. Integrating term by term: $I = \left[ \ln|u| - 3\ln|u+1| + 3\ln|u+2| - \ln|u+3| ight]_1^{\infty}$. $I = \left[ \ln \left| \frac{u(u+2)^3}{(u+1)^3(u+3)} ight| ight]_1^{\infty}$. As $u 	o \infty$,the argument of the logarithm approaches $\ln(1) = 0$. At $u = 1$,the value is $\ln \left( \frac{1(3)^3}{(2)^3(4)} ight) = \ln \left( \frac{27}{32} ight)$. Therefore,$I = 0 - \ln \left( \frac{27}{32} ight) = \ln \left( \frac{32}{27} ight)$.

$\int \limits_0^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^x+6} d x$

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