\(\displaystyle\int x\,e^{2x}\,dx\) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{e^{2x}}{4}(2x - 1) + C\)
We use integration by parts, \(\int u\,dv = uv - \int v\,du\), with
\(u = x, \quad du = dx\)
\( \displaystyle dv = e^{2x}\,dx, \quad v = \frac{e^{2x}}{2} \)
Then
\( \displaystyle \int x\,e^{2x}\,dx = \frac{x\,e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx \)
\( \displaystyle = \frac{x\,e^{2x}}{2} - \frac{e^{2x}}{4} + C \)
\( \displaystyle = \frac{e^{2x}}{4}(2x - 1) + C \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{e^{2x}}{4}(2x + 1) + C\) — Incorrect. The second term of integration by parts is subtracted, not added.
Option (3) \(\displaystyle\frac{e^{2x}}{2}(x - 1) + C\) — Incorrect. \(\int \frac{e^{2x}}{2}dx\) has been taken as \(\frac{e^{2x}}{2}\); it is \(\frac{e^{2x}}{4}\).
Option (4) \(\displaystyle\frac{x^2 e^{2x}}{4} + C\) — Incorrect. A product cannot be integrated factor by factor; the derivative of this is not \(x e^{2x}\).