Consider the following statements, where C is an arbitrary constant:
(A) \(\displaystyle\int \frac{x + 1}{x^2 + 2x + 5}\,dx = \frac{1}{2}\log(x^2 + 2x + 5) + C\)
(B) \(\displaystyle\int \frac{1}{x(x + 2)}\,dx = \log\left|\frac{x}{x + 2}\right| + C\)
(C) \(\displaystyle\int \frac{1}{2x + 3}\,dx = \log|2x + 3| + C\)
(D) \(\displaystyle\int \frac{1}{(x - 1)(x - 2)}\,dx = \log\left|\frac{x - 2}{x - 1}\right| + C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
For (A), let \(t = x^2 + 2x + 5\) (substitution):
\(dt = (2x + 2)\,dx = 2(x + 1)\,dx\)
\( \displaystyle \int \frac{x + 1}{x^2 + 2x + 5}\,dx = \frac{1}{2}\int\frac{dt}{t} \)
\( \displaystyle = \frac{1}{2}\log(x^2 + 2x + 5) + C \)
For (B) and (D), by partial fractions:
\( \displaystyle \frac{1}{x(x + 2)} = \frac{1}{2}\left(\frac{1}{x} - \frac{1}{x + 2}\right) \)
\( \displaystyle \frac{1}{(x - 1)(x - 2)} = \frac{1}{x - 2} - \frac{1}{x - 1} \)
For (C), with t = 2x + 3, dt = 2 dx:
\( \displaystyle \int \frac{dx}{2x + 3} = \frac{1}{2}\log|2x + 3| + C \)
So the four integrals are \(\frac{1}{2}\log(x^2 + 2x + 5)\), \(\frac{1}{2}\log\left|\frac{x}{x + 2}\right|\), \(\frac{1}{2}\log|2x + 3|\) and \(\log\left|\frac{x - 2}{x - 1}\right|\), each + C.
Checking the statements
(A) \(\int \frac{x + 1}{x^2 + 2x + 5}\,dx = \frac{1}{2}\log(x^2 + 2x + 5) + C\) — Correct. The numerator is half the derivative 2x + 2 of the denominator.
(B) \(\int \frac{1}{x(x + 2)}\,dx = \log\left|\frac{x}{x + 2}\right| + C\) — Incorrect. The factor \(\frac{1}{2}\) is missing; the integral is \(\frac{1}{2}\log\left|\frac{x}{x + 2}\right| + C\).
(C) \(\int \frac{1}{2x + 3}\,dx = \log|2x + 3| + C\) — Incorrect. The factor \(\frac{1}{2}\) from dt = 2 dx is missing.
(D) \(\int \frac{1}{(x - 1)(x - 2)}\,dx = \log\left|\frac{x - 2}{x - 1}\right| + C\) — Correct. \(\log|x - 2| - \log|x - 1| = \log\left|\frac{x - 2}{x - 1}\right|\).