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

Question

The answer to this question at the back of the book(from where I'm learning Integral Calculus) is given as:$ A = \frac{-3}{2}; B = \frac{35}{36}; C = \frac{-3}{2}\log{3}$

Here is my attempt at this problem:

Is my solution wrong or is C really equal to $ \frac{-3}{2}\log{3} $ somehow?

Answer 1

You made a small mistake. $$-\dfrac34\ln|z + 4| = -\dfrac34\ln|9e^{2x}| = -\dfrac34\left[\ln|e^{2x}| + \ln|9|\right] = -\dfrac32x - \dfrac32\ln3$$ From above, $C = - \dfrac32\ln3$.

Answer 2

Too long for a side comment.

To compute $A$ and $B$, we can avoid integration using $$\int{\frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}}}\,dx = Ax + B\log{(9e^{2x}-4)} + C$$ Differentiating both sides, $$\frac{4e^{2x} + 6}{9e^{2x} - 4}=A+\frac{18 e^{2 x}}{9 e^{2 x}-4}B$$ Let $e^{2x}=t$ to get $$\frac{4t + 6}{9t - 4}=A+\frac{18 t}{9 t-4}B$$ Make $t=0$ to get $A=-\frac 32$; now, make $t=1$ to get $2=A+\frac {18}5 B \implies B=\frac {35}{36}$