Question

Let $T={\int }_0^{\ln 2}\frac{3e^{3x}+2e^{2x}}{e^{3x}+e^{2x}+1}dx$, then $e^T=\frac{p}{q}$ where $p$ and $q$ are coprime to each other, then find the value of $p+q$ is

Answer 1

$T=\stackrel{10n^2}{\underset{0}{\int }}\frac{3e^{3x}+2e^{2x}}{e^{3x}+e^{2x}+1}dx$

$\Rightarrow$ Let $e^{3x}+e^{2x}+1=1,x=0,t=3$

$(3e^{3x}+2e^{2x})dx=dt,x=l{n}^2,t=13$

$T=\stackrel{13}{\underset{3}{\int }}\frac{dt}{t}$

$T=ln\left(t\right){]}_3^{13}$

$T=in\left(13\right)-in\left(3\right)$

$T=in\frac{13}{3}$ and $e^T=e{e}^{in\frac{13}{3}}$

$p=13,q=3$ $\{\therefore 9cd(13,3)=1\}$

$\Rightarrow p+q=13+3=16$