Answer
$\frac{1}{2}x^2e^{2x}- \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} +C$
Work Step by Step
Use the tabular method to find the indefinite integral.
let $u=x^2$, and $dv= e^{2x}dx$
Find all the derivates of u and the integrate dv until the derivative of u is zero.
$\int x^2e^{2x}dx$
Alternating Signs
+, -, +, -
Derivatives of u
$x^2, 2x, 2, 0$
Integrals of dv
$e^{2x}, \frac{1}{2}e^{2x}, \frac{1}{4}e^{2x}, \frac{1}{8}e^{2x}$
Using the tabular method, take the first sign, derivative of u, and the first integration of dv, Then move to the next of each value of u and dv, Repeat until the derivative is 0
$+\frac{1}{2}x^2e^{2x} + (-)\frac{1}{2}xe^{2x} +\frac{1}{4}e^{2x} +C$
