Question

If $\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+B{\log }_e\left(9e^{2x}-4\right)+C$ then $A=......,B=......,C=......$.

Answer 1

$\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$

$=\int \frac{4e^{2x}+6}{9e^{2x}-4}dx$

We can write $Numerator=A\left(Denominator\right)+B\frac{d}{dx}\left(Denominator\right)$

$4e^{2x}+6=A(9e^{2x}-4)+B\frac{d}{dx}(9e^{2x}-4)$

$4e^{2x}+6=A(9e^{2x}-4)+B\left(18e^{2x}\right)$

$4e^{2x}+6=(9A+18B)e^{2x}+(-4A)$

Comparing the coefficients, we have

$-4A=6$ or $A=-\frac{6}{4}=-\frac{3}{2}$

$9A+18B=4\Rightarrow 18B=4-9A=4+9\times \frac{3}{2}=\frac{8+27}{2}=\frac{35}{2}$

$\therefore B=\frac{35}{36}$

$\int \frac{4e^{2x}+6}{9e^{2x}-4}dx$

$=\int \frac{A(9e^{2x}-4)+B\left(18e^{2x}\right)}{9e^{2x}-4}dx$

$=\int \frac{A(9e^{2x}-4)dx}{9e^{2x}-4}+\int \frac{B\left(18e^{2x}\right)dx}{9e^{2x}-4}$

$=-\frac{3}{2}\int \frac{(9e^{2x}-4)dx}{9e^{2x}-4}+\frac{35}{36}\int \frac{\left(18e^{2x}\right)dx}{9e^{2x}-4}$

$=-\frac{3}{2}\int dx+\frac{35}{36}\int \frac{\left(18e^{2x}\right)dx}{9e^{2x}-4}$

$=-\frac{3}{2}x+\frac{35}{36}{\log }_e(9e^{2x}-4)+c$

Comparing with $Ax+B{\log }_e(9e^{2x}-4)+C$

we get $A=-\frac{3}{2},B=\frac{35}{36},C=c$