Question

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

Step2: Apply integration - by - parts

$\int x^{2}e^{7x}\mathrm{d}x=\frac{1}{7}x^{2}e^{7x}-\int\frac{1}{7}e^{7x}\cdot2x\mathrm{d}x=\frac{1}{7}x^{2}e^{7x}-\frac{2}{7}\int xe^{7x}\mathrm{d}x$.

Step3: Apply integration - by - parts again

For $\int xe^{7x}\mathrm{d}x$, let $u = x$ and $\mathrm{d}v=e^{7x}\mathrm{d}x$. Then $\mathrm{d}u=\mathrm{d}x$ and $v = \frac{1}{7}e^{7x}$. So $\int xe^{7x}\mathrm{d}x=\frac{1}{7}xe^{7x}-\int\frac{1}{7}e^{7x}\mathrm{d}x$.

Step4: Evaluate $\int\frac{1}{7}e^{7x}\mathrm{d}x$

$\int\frac{1}{7}e^{7x}\mathrm{d}x=\frac{1}{49}e^{7x}+C$. Then $\int xe^{7x}\mathrm{d}x=\frac{1}{7}xe^{7x}-\frac{1}{49}e^{7x}+C$.

Step5: Substitute back

$\int x^{2}e^{7x}\mathrm{d}x=\frac{1}{7}x^{2}e^{7x}-\frac{2}{7}(\frac{1}{7}xe^{7x}-\frac{1}{49}e^{7x})+C=\frac{1}{7}x^{2}e^{7x}-\frac{2}{49}xe^{7x}+\frac{2}{343}e^{7x}+C$.

Answer:

$\frac{1}{7}x^{2}e^{7x}-\frac{2}{49}xe^{7x}+\frac{2}{343}e^{7x}+C$