What is the integral of x^2*e^{4x} ?

The integral of x^2*e^{4x} is 1/4 e^{4x}x^2-1/32 (4e^{4x}x-e^{4x})+C

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integral of x^2*e^{4x}

\int x^{2} \cdot e^{4 x} d x

\frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}) + C

Solution steps

\int x^{2} e^{4 x} d x

Apply Integration By Parts

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

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

\int \frac{1}{2} e^{4 x} x d x = \frac{1}{32} (4 e^{4 x} x - e^{4 x})

= \frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x})

Add a constant to the solution

= \frac{1}{4} e^{4 x} x^{2} - \frac{1}{32} (4 e^{4 x} x - e^{4 x}) + C

f (x) = \frac{1}{x + 2}, at x = 9

x^{s i n (x)}

\int cos (x) cos (3 x) d x

\frac{d}{d x} (\frac{x ^{2}}{x + 1})

\frac{6 t}{6 + t}

What is the integral of x^2*e^{4x} ?
The integral of x^2*e^{4x} is 1/4 e^{4x}x^2-1/32 (4e^{4x}x-e^{4x})+C