Question

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

• A. A = -3/2, B = 35/36; C = const. of integration
• B. A = 3/2, B = -35/36; C = const. of integration
• C. A = -19/36, B = 35/36; C = const. of integration
• D. A = 19/36, B = -35/36; C = const. of integration

Answer 1

Answer: (A)

A = -3/2, B = 35/36; C = const. of integration

Answer 2

Let $I = \int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx$. We express the numerator as a linear combination of the denominator and its derivative. Let $D(x) = 9e^x - 4e^{-x}$. Then $D'(x) = 9e^x + 4e^{-x}$. We set $4e^x + 6e^{-x} = \lambda(9e^x - 4e^{-x}) + \mu(9e^x + 4e^{-x})$. Comparing coefficients of $e^x$ and $e^{-x}$: $9\lambda + 9\mu = 4$ $-4\lambda + 4\mu = 6$ Solving these equations gives $\lambda = -\frac{19}{36}$ and $\mu = \frac{35}{36}$. The integral becomes: $I = \int \left( \lambda + \mu \frac{D'(x)}{D(x)} \right) dx = \lambda x + \mu \ln|D(x)| + C$ $I = -\frac{19}{36}x + \frac{35}{36} \ln|9e^x - 4e^{-x}| + C$. Using $\ln|9e^x - 4e^{-x}| = \ln|e^{-x}(9e^{2x} - 4)| = -x + \ln|9e^{2x} - 4|$: $I = -\frac{19}{36}x + \frac{35}{36}(-x + \ln|9e^{2x} - 4|) + C$ $I = \left(-\frac{19}{36} - \frac{35}{36}\right)x + \frac{35}{36}\ln|9e^{2x} - 4| + C$ $I = -\frac{54}{36}x + \frac{35}{36}\ln|9e^{2x} - 4| + C$ $I = -\frac{3}{2}x + \frac{35}{36}\ln|9e^{2x} - 4| + C$. Comparing with $Ax+Blog(9e^{2x}-4)+C$, we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$.