# How do you find the antiderivative of x^2 e^(2x)?

$\frac{1}{2} {e}^{2 x} \left({x}^{2} - x + \frac{1}{2}\right) + C$

#### Explanation:

Well, This can't be done simply.

We have to Perform Integration By Parts.

The Rule States That,

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left(u ' \int v\right) \mathrm{dx}$

Where $u$ and $v$ are functions of $x$ and $u ' = \frac{d}{\mathrm{dx}} \left(u\right)$.

Now, We have,

$\int {x}^{2} {e}^{2 x} \mathrm{dx}$

$= {x}^{2} \int {e}^{2} x - \int \left(\frac{d}{\mathrm{dx}} \left({x}^{2}\right) \int {e}^{2 x}\right) \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \int \left(\cancel{2} x \times \frac{1}{\cancel{2}} {e}^{2 x}\right) \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \int x {e}^{2 x} \mathrm{dx}$......................(i)

We have to Integrate By Parts Once More.

So, From (i),

$\frac{1}{2} {x}^{2} {e}^{2 x} - x \int {e}^{2 x} + \int \left(\frac{d}{\mathrm{dx}} \left(x\right) \int {e}^{2 x}\right) \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \frac{1}{2} x {e}^{2 x} + \frac{1}{2} \int {e}^{2 x} \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{2 x} - \frac{1}{2} x {e}^{2 x} + \frac{1}{4} {e}^{2 x} + C$ [Yeah, Don't EVER Forget this guy here$\rightarrow$ $C$]

$= \frac{1}{2} {e}^{2 x} \left({x}^{2} - x + \frac{1}{2}\right) + C$

Hope this helps.