Question

Evaluate the definite integral: \[ \int_0^1 \frac{2x + 5}{x^2 + 3x + 2} \, dx = \]

• A. \( \log\left(\frac{16}{3}\right) \)
• B. \( 0 \)
• C. \( \log\left(\frac{3}{16}\right) \)
• D. \( 4\log2 - 2\log3 \)

Hint:

Always factor the denominator and use partial fractions when integrating rational functions. Check limits carefully after integration.

Answer 1

The Correct Option is A

We are asked to evaluate: \[ \int_0^1 \frac{2x + 5}{x^2 + 3x + 2} \, dx \] Step 1: Factor the denominator \[ x^2 + 3x + 2 = (x + 1)(x + 2) \] Step 2: Use partial fractions for the integrand Let: \[ \frac{2x + 5}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \] Multiplying both sides by \( (x + 1)(x + 2) \), we get: \[ 2x + 5 = A(x + 2) + B(x + 1) \] Expanding: \[ 2x + 5 = Ax + 2A + Bx + B = (A + B)x + (2A + B) \] Equating coefficients:
\[ \begin{aligned} A + B &= 2
\text{(i)}
2A + B &= 5
\text{(ii)} \end{aligned} \] Solving: Subtract equation (i) from equation (ii): \[ \begin{aligned} (2A + B) - (A + B) &= 5 - 2
A &= 3 \end{aligned} \] Substitute \( A = 3 \) into equation (i): \[ \begin{aligned} 3 + B &= 2
B &= -1 \end{aligned} \] Subtract (i) from (ii): \[ (2A + B) - (A + B) = 5 - 2 \Rightarrow A = 3 \Rightarrow B = -1 \] Step 3: Integrate using partial fractions \[ \int_0^1 \frac{2x + 5}{x^2 + 3x + 2} \, dx = \int_0^1 \left( \frac{3}{x + 1} - \frac{1}{x + 2} \right) dx \] \[ = \left[ 3\ln|x + 1| - \ln|x + 2| \right]_0^1 \] \[ = \left( 3\ln(2) - \ln(3) \right) - \left( 3\ln(1) - \ln(2) \right) = (3\ln2 - \ln3) - (0 - \ln2) = 3\ln2 - \ln3 + \ln2 \] \[ = 4\ln2 - \ln3 = \ln\left( \frac{16}{3} \right) \] \[ \boxed{\log\left(\frac{16}{3}\right)} \]