Consider the following statements, where C is a constant of integration:
(A) \(\int\displaystyle\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = 2e^{\sqrt{x}} + C\)
(B) \(\int\displaystyle\frac{\log x}{x}\,dx = (\log x)^2 + C\)
(C) \(\int\displaystyle\frac{4x + 3}{2x^2 + 3x + 4}\,dx = \log(2x^2 + 3x + 4) + C\)
(D) \(\int\displaystyle\frac{x}{(x^2 + 1)^2}\,dx = \frac{1}{2(x^2 + 1)} + C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
Each integral needs one substitution.
With \(t = \sqrt{x}\), \(dt = \frac{dx}{2\sqrt{x}}\):
\( \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = 2\int e^t\,dt = 2e^{\sqrt{x}} + C \)
With \(t = \log x\), \(dt = \frac{dx}{x}\):
\( \displaystyle \int \frac{\log x}{x}\,dx = \int t\,dt = \frac{(\log x)^2}{2} + C \)
With \(t = 2x^2 + 3x + 4\), \(dt = (4x + 3)\,dx\):
\( \displaystyle \int \frac{4x + 3}{2x^2 + 3x + 4}\,dx = \int \frac{dt}{t} = \log(2x^2 + 3x + 4) + C \)
No modulus is needed here because \(2x^2 + 3x + 4 = 2\left(x + \frac{3}{4}\right)^2 + \frac{23}{8} \gt 0\).
With \(t = x^2 + 1\), \(dt = 2x\,dx\):
\( \displaystyle \int \frac{x}{(x^2 + 1)^2}\,dx = \frac{1}{2}\int t^{-2}\,dt = -\frac{1}{2(x^2 + 1)} + C \)
So the four integrals are \(2e^{\sqrt{x}}\), \(\frac{1}{2}(\log x)^2\), \(\log(2x^2 + 3x + 4)\) and \(-\frac{1}{2(x^2 + 1)}\), each plus C.
Checking the statements
(A) \(\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = 2e^{\sqrt{x}} + C\) — Correct. With \(t = \sqrt{x}\) it is \(2\int e^t\,dt\).
(B) \(\int \frac{\log x}{x}\,dx = (\log x)^2 + C\) — Incorrect. The answer is \(\frac{1}{2}(\log x)^2 + C\); the factor \(\frac{1}{2}\) is missing.
(C) \(\int \frac{4x + 3}{2x^2 + 3x + 4}\,dx = \log(2x^2 + 3x + 4) + C\) — Correct. The numerator is the derivative of the denominator.
(D) \(\int \frac{x}{(x^2 + 1)^2}\,dx = \frac{1}{2(x^2 + 1)} + C\) — Incorrect. The sign is wrong: the answer is \(-\frac{1}{2(x^2 + 1)} + C\).