Match List-I with List-II (C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int x\,e^{2x}\,dx\) | (I) | \(x\,e^x + C\) |
| (B) | \(\int\displaystyle\frac{\log x}{x^2}\,dx\) | (II) | \(\displaystyle\frac{x^2}{2}\log x - \frac{x^2}{4} + C\) |
| (C) | \(\int x\log x\,dx\) | (III) | \(\displaystyle -\frac{1 + \log x}{x} + C\) |
| (D) | \(\int e^x(1 + x)\,dx\) | (IV) | \(\displaystyle\frac{1}{4}(2x - 1)e^{2x} + C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Items (A), (B) and (C) use integration by parts, \(\int u\,v\,dx = u\int v\,dx - \int \left(u' \int v\,dx\right)dx\); item (D) uses the result for \(e^x\left[f(x) + f'(x)\right]\).
(A) \(\int x\,e^{2x}\,dx\)
This matches (IV) \(\frac{1}{4}(2x - 1)e^{2x} + C\).
\( \displaystyle \int x\,e^{2x}\,dx = \frac{x\,e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx \)
\( \displaystyle = \frac{x\,e^{2x}}{2} - \frac{e^{2x}}{4} + C \)
\( \displaystyle = \frac{1}{4}(2x - 1)e^{2x} + C \)
(B) \(\int \frac{\log x}{x^2}\,dx\)
This matches (III) \(-\frac{1 + \log x}{x} + C\). With \(\log x\) as the first function and \(x^{-2}\) as the second:
\( \displaystyle \int \frac{\log x}{x^2}\,dx = -\frac{\log x}{x} - \int \frac{1}{x}\left(-\frac{1}{x}\right)dx \)
\( \displaystyle = -\frac{\log x}{x} - \frac{1}{x} + C \)
\( \displaystyle = -\frac{1 + \log x}{x} + C \)
(C) \(\int x\log x\,dx\)
This matches (II) \(\frac{x^2}{2}\log x - \frac{x^2}{4} + C\).
\( \displaystyle \int x\log x\,dx = \frac{x^2}{2}\log x - \int \frac{x^2}{2} \cdot \frac{1}{x}\,dx \)
\( \displaystyle = \frac{x^2}{2}\log x - \frac{x^2}{4} + C \)
(D) \(\int e^x(1 + x)\,dx\)
This matches (I) \(x\,e^x + C\).
With \(f(x) = x\), the bracket is \(f(x) + f'(x)\), so
\( \displaystyle \int e^x\left[f(x) + f'(x)\right]dx = x\,e^x + C \)