Question

If \( \int \frac{2e^x + 3e^{-x}}{4e^x + 7e^{-x}} dx = \frac{1}{14}(ux + v \log_e(4e^x + 7e^{-x})) + C \), where C is a constant of integration, then u + v is equal to _________.

Hint:

This problem demonstrates a standard integration pattern. Recognizing this pattern (Numerator = L\(\cdot\)Denominator + M\(\cdot\)Derivative) saves time and provides a direct path to the solution, avoiding more complex substitutions.

Answer 1

Correct Answer: 7

Step 1: Use the standard method for integrals of this form.
For an integral of the form \( \int \frac{A e^x + B e^{-x}}{C e^x + D e^{-x}} dx \), we express the numerator as a linear combination of the denominator and its derivative.
Let Numerator = \(L \cdot (\text{Denominator}) + M \cdot (\text{Derivative of Denominator})\).
- Numerator: \(N(x) = 2e^x + 3e^{-x}\)
- Denominator: \(D(x) = 4e^x + 7e^{-x}\)
- Derivative of Denominator: \(D'(x) = 4e^x - 7e^{-x}\)
Step 2: Set up and solve the system of equations for L and M.
\[ 2e^x + 3e^{-x} = L(4e^x + 7e^{-x}) + M(4e^x - 7e^{-x}) \] \[ 2e^x + 3e^{-x} = (4L+4M)e^x + (7L-7M)e^{-x} \] Comparing the coefficients of \(e^x\) and \(e^{-x}\) on both sides:
1) \(4L + 4M = 2 \implies L + M = 1/2\)
2) \(7L - 7M = 3 \implies L - M = 3/7\)
Adding (1) and (2): \(2L = \frac{1}{2} + \frac{3}{7} = \frac{7+6}{14} = \frac{13}{14} \implies L = \frac{13}{28}\).
Subtracting (2) from (1): \(2M = \frac{1}{2} - \frac{3}{7} = \frac{7-6}{14} = \frac{1}{14} \implies M = \frac{1}{28}\).
Step 3: Rewrite and evaluate the integral.
The integral becomes:
\[ \int \frac{L \cdot D(x) + M \cdot D'(x)}{D(x)} dx = \int \left( L + M \frac{D'(x)}{D(x)} \right) dx \] \[ = L \int dx + M \int \frac{D'(x)}{D(x)} dx = Lx + M \ln|D(x)| + C \] Substituting L and M:
\[ \frac{13}{28}x + \frac{1}{28} \ln(4e^x + 7e^{-x}) + C \] Step 4: Compare with the given form to find u and v.
The given form is \(\frac{1}{14}(ux + v \log_e(4e^x + 7e^{-x})) + C\).
Let's factor out \(\frac{1}{14}\) from our result:
\[ \frac{1}{14} \left( \frac{14 \cdot 13}{28}x + \frac{14 \cdot 1}{28} \ln(4e^x + 7e^{-x}) \right) + C \] \[ \frac{1}{14} \left( \frac{13}{2}x + \frac{1}{2} \ln(4e^x + 7e^{-x}) \right) + C \] By comparing this with the given form, we identify:
- \(u = \frac{13}{2}\)
- \(v = \frac{1}{2}\)
Step 5: Calculate u + v.
\[ u + v = \frac{13}{2} + \frac{1}{2} = \frac{14}{2} = 7 \]