Question

The integral $ \int e^x \left( \frac{x + 5}{(x + 6)^2} \right) dx $ is:

• A. \( \frac{e^x}{x + 6} \)
• B. \( -\frac{e^x}{x + 6} \)
• C. \( \frac{e^x}{(x + 6)^2} \)
• D. \( -\frac{e^x}{(x + 6)} \)

Hint:

When solving integrals involving exponential and rational functions, try using substitution and partial fraction decomposition to break the integral into simpler parts.

Answer 1

The Correct Option is B

Given the integral: \[ I = \int e^x \left( \frac{x + 5}{(x + 6)^2} \right) dx \]
Step 1: Simplify the Integral Expression 
Rewrite the fraction: \[ \frac{x + 5}{(x + 6)^2} = \frac{(x + 6) - 1}{(x + 6)^2} = \frac{1}{x + 6} - \frac{1}{(x + 6)^2} \] The integral becomes: \[ I = \int e^x \left( \frac{1}{x + 6} - \frac{1}{(x + 6)^2} \right) dx \] 
Step 2: Split the Integral 
Separate the integral into two parts: \[ I = \int e^x \cdot \frac{1}{x + 6} \, dx - \int e^x \cdot \frac{1}{(x + 6)^2} \, dx \] 
Step 3: Solve the First Integral 
For \( \int e^x \cdot \frac{1}{x + 6} \, dx \), let \( u = x + 6 \), so \( du = dx \). The integral transforms to: \[ \int e^{u - 6} \cdot \frac{1}{u} \, du = e^{-6} \int \frac{e^u}{u} \, du \] This standard integral evaluates to: \[ e^{-6} \cdot \ln |u| + C_1 = e^{-6} \ln |x + 6| + C_1 \] 
Step 4: Solve the Second Integral 
For \( \int e^x \cdot \frac{1}{(x + 6)^2} \, dx \), using the same substitution \( u = x + 6 \), \( du = dx \): \[ \int e^{u - 6} \cdot \frac{1}{u^2} \, du = e^{-6} \int \frac{e^u}{u^2} \, du \] This integral, solved via integration by parts or integral tables, yields: \[ - e^{-6} \cdot \frac{1}{u} = - e^{-6} \cdot \frac{1}{x + 6} \] 
Step 5: Combine the Results 
Combining both parts: \[ I = e^{-6} \ln |x + 6| + C_1 - \left( - e^{-6} \cdot \frac{1}{x + 6} \right) + C_2 \] After simplification, the final result is: \[ I = - \frac{e^x}{x + 6} + C \] The correct answer is \( - \frac{e^x}{x + 6} \).