Question

If \(\displaystyle \int \frac{3x+2}{4x^2+4x+5} \, dx = A \log(4x^2+4x+5) + B \tan^{-1} \left(x + \frac{1}{2}\right) + C\), then \((A, B) =\)

• A. \(\left(\frac{3}{8}, \frac{1}{8}\right)\)
• B. \(\left(\frac{5}{8}, \frac{1}{8}\right)\)
• C. \(\left(-\frac{3}{8}, \frac{1}{8}\right)\)
• D. \(\left(-\frac{5}{8}, \frac{1}{8}\right)\)

Hint:

For rational functions with irreducible quadratics in the denominator, try completing the square and matching with standard integrals.

Answer 1

The Correct Option is A

Let us write the integrand as: \[ \frac{3x+2}{4x^2+4x+5} \] We observe the denominator is quadratic. Try substitution or decomposition. Complete the square in the denominator: \[ 4x^2+4x+5 = 4(x^2+x) + 5 = 4\left(x+\frac{1}{2}\right)^2 + 4 \] Now, use the standard form: \[ \int \frac{ax + b}{(x + c)^2 + d^2} dx = A \log((x + c)^2 + d^2) + B \tan^{-1} \left( \frac{x + c}{d} \right) \] Match to get: \[ A = \frac{3}{8}, \quad B = \frac{1}{8} \]