What is the integral of x^2e^{4x+3} ?

The integral of x^2e^{4x+3} is 1/4 e^{4x+3}x^2-1/2 (1/4 e^{4x+3}x-1/16 e^{4x+3})+C

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integral of x^2e^{4x+3}

\int x^{2} e^{4 x + 3} d x

\frac{1}{4} e^{4 x + 3} x^{2} - \frac{1}{2} (\frac{1}{4} e^{4 x + 3} x - \frac{1}{16} e^{4 x + 3}) + C

Solution steps

\int x^{2} e^{4 x + 3} d x

Apply Integration By Parts

: \frac{1}{4} e^{4 x + 3} x^{2} - \int \frac{1}{2} e^{4 x + 3} x d x

= \frac{1}{4} e^{4 x + 3} x^{2} - \int \frac{1}{2} e^{4 x + 3} x d x

\int \frac{1}{2} e^{4 x + 3} x d x = \frac{1}{2} (\frac{1}{4} e^{4 x + 3} x - \frac{1}{16} e^{4 x + 3})

= \frac{1}{4} e^{4 x + 3} x^{2} - \frac{1}{2} (\frac{1}{4} e^{4 x + 3} x - \frac{1}{16} e^{4 x + 3})

Add a constant to the solution

= \frac{1}{4} e^{4 x + 3} x^{2} - \frac{1}{2} (\frac{1}{4} e^{4 x + 3} x - \frac{1}{16} e^{4 x + 3}) + C

\int \frac{( x + 6 )}{5 - 4 x - x ^{2}} d x

9 - x

f (x) = (\frac{4 x}{7 x + 5})^{4}

y^{^{''}} + y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = 1

x \cdot \frac{d y}{d x} = y^{2} - y

What is the integral of x^2e^{4x+3} ?
The integral of x^2e^{4x+3} is 1/4 e^{4x+3}x^2-1/2 (1/4 e^{4x+3}x-1/16 e^{4x+3})+C