_0^6e^3x+6e^2x+11e^x+6dx=

Question

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

Accepted Answer

$\begin{gathered} l={\int }_{\infty}^{0}\frac{6}{({e}^{x}+1)({e}^{x}+2)({e}^{x}+3)}dx \ =6{\int }_{\infty}^{0}(\frac{\frac{1}{2}}{{e}^{x}+1}+\frac{-1}{{e}^{x}+2}+\frac{\frac{1}{2}}{{e}^{x}+3})dx \ =3{\int }_{\infty}^{0}\frac{{e}^{-x}}{1+{e}^{-x}}dx-6{\int }_{\infty}^{0}\frac{{e}^{-x}dx}{1+2{e}^{-x}}+3{\int }_{\infty}^{0}\frac{{e}^{-x}}{1+3{e}^{-x}}dx \ =3{[-\ln (1+{e}^{-x})]}_{\infty}^{0}+6\frac{1}{2}{[\ln (1+2{e}^{-x})]}_{\infty}^{0}-\frac{3}{3}{[\ln (1+3{e}^{-x})]}_{\infty}^{0} \ =3\ln 2-3\ln 3+\ln 4 \ =3\ln \frac{2}{3}+\ln 4 \ =\ln \frac{32}{27} \end{gathered}$

Solutions · 1