evaluate the integral using integration by parts.\n int(9x^{2}-8x)e^{2x}dx \n int(9x^{2}-8x)e^{2x}dx=square

Explanation:

Step1: Recall integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = 9x^{2}-8x$ and $\mathrm{d}v=e^{2x}\mathrm{d}x$. Then $\mathrm{d}u=(18x - 8)\mathrm{d}x$ and $v=\frac{1}{2}e^{2x}$.

Step2: Apply integration - by - parts

$\int(9x^{2}-8x)e^{2x}\mathrm{d}x=(9x^{2}-8x)\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}(18x - 8)\mathrm{d}x$ $=\frac{1}{2}(9x^{2}-8x)e^{2x}-\frac{1}{2}\int(18x - 8)e^{2x}\mathrm{d}x$

Step3: Apply integration - by - parts again

For $\int(18x - 8)e^{2x}\mathrm{d}x$, let $u = 18x-8$, $\mathrm{d}v=e^{2x}\mathrm{d}x$. Then $\mathrm{d}u = 18\mathrm{d}x$ and $v=\frac{1}{2}e^{2x}$. $\int(18x - 8)e^{2x}\mathrm{d}x=(18x - 8)\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}\times18\mathrm{d}x$ $=(9x - 4)e^{2x}-9\int e^{2x}\mathrm{d}x$

Step4: Integrate $e^{2x}$

Since $\int e^{2x}\mathrm{d}x=\frac{1}{2}e^{2x}+C$, then $\int(18x - 8)e^{2x}\mathrm{d}x=(9x - 4)e^{2x}-\frac{9}{2}e^{2x}+C$ $=(9x-4-\frac{9}{2})e^{2x}+C=(9x-\frac{8 + 9}{2})e^{2x}+C=(9x-\frac{17}{2})e^{2x}+C$

Step5: Substitute back

$\int(9x^{2}-8x)e^{2x}\mathrm{d}x=\frac{1}{2}(9x^{2}-8x)e^{2x}-\frac{1}{2}(9x-\frac{17}{2})e^{2x}+C$ $=\left(\frac{9x^{2}}{2}-4x-\frac{9x}{2}+\frac{17}{4}\right)e^{2x}+C$ $=\left(\frac{9x^{2}}{2}-\frac{8x + 9x}{2}+\frac{17}{4}\right)e^{2x}+C$ $=\left(\frac{9x^{2}}{2}-\frac{17x}{2}+\frac{17}{4}\right)e^{2x}+C$

Answer:

$\left(\frac{9x^{2}}{2}-\frac{17x}{2}+\frac{17}{4}\right)e^{2x}+C$