\(\displaystyle\int x^3 e^{x^2}\,dx\) is equal to (where c is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\left(x^2 - 1\right)e^{x^2} + c\)
Let \(t = x^2\). Then \(dt = 2x\,dx\), so \(x\,dx = \frac{dt}{2}\), and \(x^3\,dx = x^2 \cdot x\,dx = \frac{t}{2}\,dt\). By this substitution:
\( \displaystyle \int x^3 e^{x^2}\,dx = \frac{1}{2}\int t\,e^{t}\,dt \)
By integration by parts, \(\int t\,e^{t}\,dt = t\,e^{t} - e^{t} = (t - 1)e^{t}\). So
\( \displaystyle \int x^3 e^{x^2}\,dx = \frac{1}{2}(t - 1)e^{t} + c \)
\( \displaystyle = \frac{1}{2}\left(x^2 - 1\right)e^{x^2} + c \)
Check: the derivative of \(\frac{1}{2}(x^2 - 1)e^{x^2}\) is \(x\,e^{x^2} + \frac{1}{2}(x^2 - 1) \cdot 2x\,e^{x^2} = x^3 e^{x^2}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\left(x^2 + 1\right)e^{x^2} + c\) — Incorrect. The sign in \(\int t\,e^{t}\,dt = (t - 1)e^{t}\) has been taken as plus; the derivative is \((x^3 + 2x)e^{x^2}\).
Option (2) \(\left(x^2 - 1\right)e^{x^2} + c\) — Incorrect. The factor \(\frac{1}{2}\) from \(x\,dx = \frac{dt}{2}\) has been lost; the derivative is \(2x^3 e^{x^2}\).
Option (3) \(\displaystyle\frac{1}{2}x^2 e^{x^2} + c\) — Incorrect. This stops after the first term of integration by parts; the derivative is \((x + x^3)e^{x^2}\).