\(\displaystyle\int \frac{3x + 5}{(x + 1)(x + 3)}\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\log|x + 1| + 2\log|x + 3| + C\)
By partial fractions, let
\( \displaystyle \frac{3x + 5}{(x + 1)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3} \)
\(3x + 5 = A(x + 3) + B(x + 1)\)
Putting x = −1 and then x = −3:
\(2 = 2A \ \Rightarrow \ A = 1\)
\(-4 = -2B \ \Rightarrow \ B = 2\)
So
\( \displaystyle \int \frac{3x + 5}{(x + 1)(x + 3)}\,dx = \int \frac{dx}{x + 1} + \int \frac{2\,dx}{x + 3} \)
\(= \log|x + 1| + 2\log|x + 3| + C\)
Why the other options are incorrect
Option (1) \(2\log|x + 1| + \log|x + 3| + C\) — Incorrect. A and B have been swapped; this differentiates to \(\frac{3x + 7}{(x + 1)(x + 3)}\).
Option (3) \(\log|x + 1| - 2\log|x + 3| + C\) — Incorrect. B = +2, not −2: at x = −3 both sides of \(-4 = -2B\) are negative.
Option (4) \(\displaystyle\frac{3}{2}\log|(x + 1)(x + 3)| + C\) — Incorrect. This gives A = B = 3/2, which differentiates to \(\frac{3x + 6}{(x + 1)(x + 3)}\).