Question

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

Step2: Find $\mathrm{d}u$ and $v$

Differentiate $u = 2x$ with respect to $x$ to get $\mathrm{d}u=2\mathrm{d}x$. Integrate $\mathrm{d}v = e^{7x}\mathrm{d}x$ with respect to $x$. Using the formula $\int e^{ax}\mathrm{d}x=\frac{1}{a}e^{ax}+C$ ($a = 7$ here), we have $v=\frac{1}{7}e^{7x}$.

Step3: Apply the integration - by - parts formula

$\int 2xe^{7x}\mathrm{d}x=2x\cdot\frac{1}{7}e^{7x}-\int\frac{1}{7}e^{7x}\cdot2\mathrm{d}x$.

Step4: Evaluate the remaining integral

$\int\frac{2}{7}e^{7x}\mathrm{d}x=\frac{2}{7}\cdot\frac{1}{7}e^{7x}+C=\frac{2}{49}e^{7x}+C$.
So, $\int 2xe^{7x}\mathrm{d}x=\frac{2}{7}xe^{7x}-\frac{2}{49}e^{7x}+C$.

Answer:

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