What is the integral of (3-4x^2)e^{8x-5} ?

The integral of (3-4x^2)e^{8x-5} is -1/2 e^{8x-5}x^2+1/8 e^{8x-5}x+23/64 e^{8x-5}+C

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integral of (3-4x^2)e^{8x-5}

\int (3 - 4 x^{2}) e^{8 x - 5} d x

- \frac{1}{2} e^{8 x - 5} x^{2} + \frac{1}{8} e^{8 x - 5} x + \frac{23}{64} e^{8 x - 5} + C

Solution steps

\int (3 - 4 x^{2}) e^{8 x - 5} d x

Apply Integration By Parts

= \frac{1}{8} e^{8 x - 5} (3 - 4 x^{2}) - \int - e^{8 x - 5} x d x

\int - e^{8 x - 5} x d x = - \frac{1}{8} e^{8 x - 5} x + \frac{1}{64} e^{8 x - 5}

= \frac{1}{8} e^{8 x - 5} (3 - 4 x^{2}) - (- \frac{1}{8} e^{8 x - 5} x + \frac{1}{64} e^{8 x - 5})

Simplify

\frac{1}{8} e^{8 x - 5} (3 - 4 x^{2}) - (- \frac{1}{8} e^{8 x - 5} x + \frac{1}{64} e^{8 x - 5}) : - \frac{1}{2} e^{8 x - 5} x^{2} + \frac{1}{8} e^{8 x - 5} x + \frac{23}{64} e^{8 x - 5}

= - \frac{1}{2} e^{8 x - 5} x^{2} + \frac{1}{8} e^{8 x - 5} x + \frac{23}{64} e^{8 x - 5}

Add a constant to the solution

= - \frac{1}{2} e^{8 x - 5} x^{2} + \frac{1}{8} e^{8 x - 5} x + \frac{23}{64} e^{8 x - 5} + C

What is the integral of (3-4x^2)e^{8x-5} ?
The integral of (3-4x^2)e^{8x-5} is -1/2 e^{8x-5}x^2+1/8 e^{8x-5}x+23/64 e^{8x-5}+C