Question

$\begin{array}{c}\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B{\log }_e\left(9e^{2x}-4\right)C\\ A=\_\_\_\_\_,B=\_\_\_\_\_,C=\_\_\_\_\_\end{array}$

Answer 1

Lets write

$4e^x+6e^{-x}=A(9e^X-4e^{-x})+B\frac{d(9e^x-4e^{-x})}{dx}$

$=A(9e^x-4e^{-x})+B(9e^x+4e^{-x})$

$e^e(9A+9B)+(-4A+4B)e^{-x}$

by comparing both sides, we get

$9A+9B=4$ of $A+B=\frac{4}{9}\therefore B=\frac{4}{9}-A$

and $-4A+4B=6$

$-4A+4(\frac{4}{9}-A)=6$

$-4A+\frac{16}{9}-4A=6$

$-8A=6-\frac{16}{9}$

$-8A=\frac{34-16}{9}$

$-8A=\frac{38}{9}$

$\therefore A=\frac{-38}{8\times 9}=\frac{-19}{36}$

$\therefore B=\frac{4}{9}-\left(\frac{-19}{36}\right)$

$=\frac{16+19}{36}=\frac{35}{36}$

$\therefore \int \frac{\frac{-19}{36}({19}^x-4e^{-x})+\frac{35}{36}({19}^x+4e^{-x})}{(9e^x-4e^{-x})}$

$=\frac{-19}{36}\int dx+\frac{35}{36}\int \frac{{19}^x+4e^{-x}}{{19}^x-4e^{-x}}dx$

let ${19}^x-4e^{-x}=t$

$=\therefore ({19}^x-4e^{-x})dx=dt$

$=\therefore I=\frac{-19}{36}x+\frac{35}{36}\log ({19}^{2x}-4e^{-x})+c$

$=\frac{-19}{36}x+\frac{35}{36}\log \left(\frac{{19}^{2x}-4}{e^{2x}}\right)+c$

$=\frac{-19}{36}x+\frac{35}{36}\left[\log \right({19}^{2x}-4)-\log e^{2x}]+c$

$=\frac{-19}{36}x+\frac{35}{36}\log ({19}^{2x}-4)-\frac{35}{36}\times 2x+c$

$=\frac{-89}{36}x+\frac{35}{36}\log ({19}^{2x}-4)+c$