\(\int\displaystyle\frac{x}{e^{x^2}}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle -\frac{1}{2}e^{-x^2} + C\)
Since \(\frac{1}{e^{x^2}} = e^{-x^2}\), the integral is \(\int x\,e^{-x^2}\,dx\).
Let \(t = -x^2\). Then
\(dt = -2x\,dx\)
\(x\,dx = -\frac{1}{2}\,dt\)
\( \displaystyle \int x\,e^{-x^2}\,dx = -\frac{1}{2}\int e^{t}\,dt \)
\( \displaystyle = -\frac{1}{2}e^{-x^2} + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}e^{-x^2} + C\) — Incorrect. The minus sign from \(dt = -2x\,dx\) is lost; its derivative is \(-x\,e^{-x^2}\).
Option (2) \(\displaystyle -\frac{1}{2}e^{x^2} + C\) — Incorrect. Its derivative is \(-x\,e^{x^2}\), which has \(e^{x^2}\) in the numerator instead of the denominator.
Option (3) \(-2e^{-x^2} + C\) — Incorrect. Its derivative is \(4x\,e^{-x^2}\); the factor should be \(-\frac{1}{2}\), not \(-2\).