If 2e^x+3e^-x4e^x+7e^-x,dx=114(ux+v_e(4e^x+7e^-x))+C, where C is a constant of integration…

Question

If $\displaystyle\int\frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}\,dx=\dfrac{1}{14}\left(ux+v\log_e(4e^x+7e^{-x})\right)+C$, where C is a constant of integration, then $u+v$ is equal to

Answer: 7

Accepted Answer

$$I=\int\frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}dx=\int\frac{2e^{2x}+3}{4e^{2x}+7}dx$$

Write $2e^{2x}+3=A\cdot\dfrac{d}{dx}(4e^{2x}+7)+B(4e^{2x}+7)$: $2e^{2x}+3=(8A+4B)e^{2x}+7B$, giving $B=\dfrac37,\ A=\dfrac1{28}$.
$$I=\frac1{28}\ln(4e^{2x}+7)+\frac37x+C=\frac1{28}\left[x+\ln(4e^x+7e^{-x})\right]+\frac37x+C=\frac1{14}\left(\frac{13}{2}x+\frac12\ln(4e^x+7e^{-x})\right)+C$$

So $u=\dfrac{13}2,\ v=\dfrac12\Rightarrow u+v=7$

Solutions · 1