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


ANSWER

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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