$\displaystyle\int heta^2 e^{2 heta}\,d heta$ equals:

Question

$\displaystyle\int	heta^2 e^{2	heta}\,d	heta$ equals:

Accepted Answer

D. $\frac{1}{4}e^{2	heta}(2	heta^2-2	heta+1)+c$

1. Use integration by parts twice, with $\int u\,dv=uv-\int v\,du$.
2. First pass: $u=	heta^2$, $dv=e^{2	heta}d	heta$ gives $\dfrac{	heta^2e^{2	heta}}{2}-\int	heta e^{2	heta}d	heta$.
3. Second pass on $\int	heta e^{2	heta}d	heta$ gives $\dfrac{	heta e^{2	heta}}{2}-\dfrac{e^{2	heta}}{4}$.
4. Combining: $\dfrac{	heta^2e^{2	heta}}{2}-\dfrac{	heta e^{2	heta}}{2}+\dfrac{e^{2	heta}}{4}+c$.
5. Factoring $\dfrac{e^{2	heta}}{4}$ gives $\dfrac{1}{4}e^{2	heta}(2	heta^2-2	heta+1)+c$. The middle sign is minus; a plus there means one parts step dropped its sign.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_

Answer

$\frac{1}{4}e^{2	heta}(2	heta^2+2	heta+1)+c$

Answer

$\frac{1}{4}e^{2	heta}(	heta^2-2	heta+1)+c$

Answer

$\frac{1}{2}e^{2	heta}(	heta^2-	heta)+c$

$\displaystyle\int\theta^2 e^{2\theta}\,d\theta$ equals: