Question

Evaluate the integral \( \int_0^5 \frac{dx}{x^2 + 2x + 10} \)

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{12} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{4} \)

Hint:

To solve integrals of the form \( \int \frac{dx}{x^2 + a^2} \), use the arctangent formula: \( \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \).

Answer 1

The Correct Option is B

Step 1: Complete the square in the denominator.
The denominator is \( x^2 + 2x + 10 \). To complete the square, we write: \[ x^2 + 2x + 10 = (x + 1)^2 + 9 \]
Step 2: Use the standard formula for integrals.
Now, the integral becomes: \[ \int_0^5 \frac{dx}{(x+1)^2 + 3^2} \] This is a standard arctangent integral of the form \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \).
Step 3: Evaluate the integral.
Applying the formula, we get: \[ \frac{1}{3} \left[ \tan^{-1} \left( \frac{x+1}{3} \right) \right]_0^5 \] Substituting the limits: \[ = \frac{1}{3} \left( \tan^{-1} \left( \frac{6}{3} \right) - \tan^{-1} \left( \frac{1}{3} \right) \right) = \frac{1}{3} \left( \frac{\pi}{3} - \frac{1}{3} \right) = \frac{\pi}{12} \]
Step 4: Conclusion.
Thus, the value of the integral is \( \frac{\pi}{12} \).