evaluate the integral using integration by parts.\n int (7x^{2}-4x)e^{2x}dx \n int (7x^{2}-4x)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 $u = 7x^{2}-4x$ and $\mathrm{d}v=e^{2x}\mathrm{d}x$. Then $\mathrm{d}u=(14x - 4)\mathrm{d}x$ and $v=\frac{1}{2}e^{2x}$.

Step2: Apply integration - by - parts once

$\int(7x^{2}-4x)e^{2x}\mathrm{d}x=(7x^{2}-4x)\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}(14x - 4)\mathrm{d}x=\frac{1}{2}(7x^{2}-4x)e^{2x}-\frac{1}{2}\int(14x - 4)e^{2x}\mathrm{d}x$.

Step3: Apply integration - by - parts again

For $\int(14x - 4)e^{2x}\mathrm{d}x$, let $u = 14x-4$ and $\mathrm{d}v=e^{2x}\mathrm{d}x$. Then $\mathrm{d}u = 14\mathrm{d}x$ and $v=\frac{1}{2}e^{2x}$. $\int(14x - 4)e^{2x}\mathrm{d}x=(14x - 4)\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}\times14\mathrm{d}x=(7x - 2)e^{2x}-7\int e^{2x}\mathrm{d}x$.

Step4: Integrate $e^{2x}$

Since $\int e^{2x}\mathrm{d}x=\frac{1}{2}e^{2x}+C$, then $\int(14x - 4)e^{2x}\mathrm{d}x=(7x - 2)e^{2x}-\frac{7}{2}e^{2x}+C=(7x-2-\frac{7}{2})e^{2x}+C=(7x-\frac{4 + 7}{2})e^{2x}+C=(7x-\frac{11}{2})e^{2x}+C$.

Step5: Substitute back

$\int(7x^{2}-4x)e^{2x}\mathrm{d}x=\frac{1}{2}(7x^{2}-4x)e^{2x}-\frac{1}{2}(7x-\frac{11}{2})e^{2x}+C$ $=\left(\frac{7x^{2}}{2}-2x-\frac{7x}{2}+\frac{11}{4}\right)e^{2x}+C=\left(\frac{7x^{2}}{2}-\frac{7x + 4x}{2}+\frac{11}{4}\right)e^{2x}+C=\left(\frac{7x^{2}}{2}-\frac{11x}{2}+\frac{11}{4}\right)e^{2x}+C$.

Answer:

$\left(\frac{7x^{2}}{2}-\frac{11x}{2}+\frac{11}{4}\right)e^{2x}+C$