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

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

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

\int + x^{2} 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

y^{^{''}} + 16 y = e^{- t}

\frac{\partial}{\partial x} (e^{- x} cos (x))

9 y ln (x) - x y^{^{'}} = 0

36 y^{^{''}} + 84 y^{^{'}} + 49 y = 0

(x sin (3 x))^{^{'}}

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