What is the integral of x^2e^{2x+1} ?

The integral of x^2e^{2x+1} is 1/2 e^{2x+1}x^2-1/2 e^{2x+1}x+1/4 e^{2x+1}+C

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integral of x^2e^{2x+1}

\int x^{2} e^{2 x + 1} d x

\frac{1}{2} e^{2 x + 1} x^{2} - \frac{1}{2} e^{2 x + 1} x + \frac{1}{4} e^{2 x + 1} + C

Solution steps

\int x^{2} e^{2 x + 1} d x

Apply Integration By Parts

: \frac{1}{2} e^{2 x + 1} x^{2} - \int e^{2 x + 1} x d x

= \frac{1}{2} e^{2 x + 1} x^{2} - \int e^{2 x + 1} x d x

\int e^{2 x + 1} x d x = \frac{1}{2} e^{2 x + 1} x - \frac{1}{4} e^{2 x + 1}

= \frac{1}{2} e^{2 x + 1} x^{2} - (\frac{1}{2} e^{2 x + 1} x - \frac{1}{4} e^{2 x + 1})

Simplify

= \frac{1}{2} e^{2 x + 1} x^{2} - \frac{1}{2} e^{2 x + 1} x + \frac{1}{4} e^{2 x + 1}

Add a constant to the solution

= \frac{1}{2} e^{2 x + 1} x^{2} - \frac{1}{2} e^{2 x + 1} x + \frac{1}{4} e^{2 x + 1} + C

f (x) = - 3 x^{2} - 5, at x = 0

\frac{d}{d x} (\frac{x - 1}{x - 2})

\int \frac{tan ( θ )}{sec ^{2} ( θ )} d θ

\frac{8}{x}

\frac{d}{d x} (a (1 - cos^{2} (\frac{x}{2}))^{2})

What is the integral of x^2e^{2x+1} ?
The integral of x^2e^{2x+1} is 1/2 e^{2x+1}x^2-1/2 e^{2x+1}x+1/4 e^{2x+1}+C