Mathematics

# Let $T= \int_0^{\ln2} \frac{3e^{3x} + 2e^{2x} }{e^{3x}+e^{2x} +1} dx$, then $e^T=\frac{p}{q}$ where $p$ and $q$ are coprime to each other, then find the value of $p+q$ is

16

##### SOLUTION
$T=\overset{10n^2}{\underset{0}{\int}}\dfrac{3e^{3x}+2e^{2x}}{e^{3x}+e^{2x}+1}dx$
$\Rightarrow$ Let $e^{3x}+e^{2x}+1=1,x=0,t=3$
$(3e^{3x}+2e^{2x})dx=dt,x=ln^2,t=13$
$T=\overset{13}{\underset{3}{\int}}\dfrac{dt}{t}$
$T=ln(t)]^{13}_3$
$T=in(13)-in(3)$
$T=in\dfrac{13}{3}$   and $e^T=ee^{in\dfrac{13}{3}}$
$p=13,q=3$   $\{ \therefore 9cd(13,3)=1\}$
$\Rightarrow \boxed {p+q=13+3=16}$

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