Question

If $\underset{0}{\overset{\ln 4}{\int }}\frac{3e^{4x}+2e^{3x}+e^{2x}-4}{e^{4x}+e^{3x}+e^{2x}+e^x+4}dx=\ln \left(\frac{a}{4}\right)$, where $a\in R$, then the value of $a$ is

Answer 1

$I=\underset{0}{\overset{\ln 4}{\int }}\frac{3e^{4x}+2e^{3x}+e^{2x}-4}{e^{4x}+e^{3x}+e^{2x}+e^x+4}dx\Rightarrow I=\underset{0}{\overset{\ln 4}{\int }}[\frac{4e^{4x}+3e^{3x}+2e^{2x}+e^x}{e^{4x}+e^{3x}+e^{2x}+e^x+4}-\frac{e^{4x}+e^{3x}+e^{2x}+e^x+4}{e^{4x}+e^{3x}+e^{2x}+e^x+4}]dx$
$\Rightarrow I=\underset{0}{\overset{\ln 4}{\int }}[\frac{4e^{4x}+3e^{3x}+2e^{2x}+e^x}{e^{4x}+e^{3x}+e^{2x}+e^x+4}-1]dx$
$\Rightarrow I={\left[\ln \right(e^{4x}+e^{3x}+e^{2x}+e^x+4)-x]}_0^{\ln 4}\Rightarrow I=\ln (4^4+4^3+4^2+4^1+4)-\ln \left(8\right)-\ln \left(4\right)\Rightarrow I=\ln \left(\frac{43}{4}\right)\therefore a=43$