Hello,
∫ x³ e^(x²) dx =
make a substitution, first; rewrite the integral as:
∫ x² e^(x²) x dx =
let:
x² = t
differentiate both sides:
d(x²) = dt
2x dx = dt
x dx = (1/2) dt
then, substituting:
∫ x² e^(x²) x dx = ∫ t e^t (1/2) dt =
integrate this by parts, letting:
e^t dt = dv → e^t = v
(1/2)t = u → (1/2) dt = du
yielding:
∫ u dv = u v - ∫ v du
∫ (1/2)t e^t dt = (1/2)t e^t - ∫ e^t (1/2) dt =
(1/2)t e^t - ∫ (1/2) e^t dt =
(1/2)t e^t - (1/2)e^t + C =
(1/2)e^t (t - 1) + C
substitute back x² for u, ending with:
∫ x³ e^(x²) dx = (1/2)e^(x²) (x² - 1) + C
I hope it's helpful
∫ x³ e^(x²) dx =
make a substitution, first; rewrite the integral as:
∫ x² e^(x²) x dx =
let:
x² = t
differentiate both sides:
d(x²) = dt
2x dx = dt
x dx = (1/2) dt
then, substituting:
∫ x² e^(x²) x dx = ∫ t e^t (1/2) dt =
integrate this by parts, letting:
e^t dt = dv → e^t = v
(1/2)t = u → (1/2) dt = du
yielding:
∫ u dv = u v - ∫ v du
∫ (1/2)t e^t dt = (1/2)t e^t - ∫ e^t (1/2) dt =
(1/2)t e^t - ∫ (1/2) e^t dt =
(1/2)t e^t - (1/2)e^t + C =
(1/2)e^t (t - 1) + C
substitute back x² for u, ending with:
∫ x³ e^(x²) dx = (1/2)e^(x²) (x² - 1) + C
I hope it's helpful