Question

$\int \frac{3e^x + 5e^{-x}}{4e^x - 5e^{-x}}dx = Ax + B \ln |4e^{2x} - 5| + C$ the

• A. A=-1, B=-7/8; C = const. of integration
• B. A = 1, B = 7/8; C = const. of integration
• C. A=-1/8, B = 7/8; C = const. of integration
• D. A=-1, B=7/8; C = const. of integration

Answer 1

Answer: (D)

(4)

Answer 2

To solve the integral $I = \int \frac{3e^x + 5e^{-x}}{4e^x - 5e^{-x}}dx$, we express the numerator as a linear combination of the denominator and its derivative.

Let $D(x) = 4e^x - 5e^{-x}$, then $D'(x) = 4e^x + 5e^{-x}$.

We want to find $\lambda$ and $\mu$ such that $3e^x + 5e^{-x} = \lambda (4e^x - 5e^{-x}) + \mu (4e^x + 5e^{-x})$. This gives us the equations:

$4\lambda + 4\mu = 3$ and $-5\lambda + 5\mu = 5$.

Solving this system, we find $\lambda = -1/8$ and $\mu = 7/8$.

Thus, $I = \int \frac{-\frac{1}{8}(4e^x - 5e^{-x}) + \frac{7}{8}(4e^x + 5e^{-x})}{4e^x - 5e^{-x}}dx = -\frac{1}{8} \int dx + \frac{7}{8} \int \frac{4e^x + 5e^{-x}}{4e^x - 5e^{-x}} dx$.

The second integral is $\frac{7}{8} \ln|4e^x - 5e^{-x}|$. So, $I = -\frac{1}{8}x + \frac{7}{8} \ln|4e^x - 5e^{-x}| + C$.

Since $\ln|4e^x - 5e^{-x}| = \ln|e^{-x}(4e^{2x} - 5)| = -x + \ln|4e^{2x} - 5|$, we have

$I = -\frac{1}{8}x + \frac{7}{8}(-x + \ln|4e^{2x} - 5|) + C = -x + \frac{7}{8} \ln|4e^{2x} - 5| + C$.

Comparing with $Ax + B \ln |4e^{2x} - 5| + C$, we get $A = -1$ and $B = 7/8$.