$\\int (4x - 7)e^{-7x}dx=$ (use $c$ for the constant of antidifferentiation)

Explanation:

Step1: Apply integration by parts formula $\int u dv=uv-\int v du$

Let $u = 4x - 7$, $dv=e^{-7x}dx$. Then $du = 4dx$ and $v=-\frac{1}{7}e^{-7x}$.

Step2: Substitute into the formula

$\int(4x - 7)e^{-7x}dx=(4x - 7)\left(-\frac{1}{7}e^{-7x}\right)-\int\left(-\frac{1}{7}e^{-7x}\right)\times4dx$.

Step3: Simplify the first - term and solve the second integral

The first - term is $-\frac{4x - 7}{7}e^{-7x}$. The second integral: $\int\left(-\frac{4}{7}e^{-7x}\right)dx=\frac{4}{49}e^{-7x}+c$.

Step4: Combine the terms

$-\frac{4x - 7}{7}e^{-7x}-\frac{4}{49}e^{-7x}+c=-\frac{4x}{7}e^{-7x}+e^{-7x}-\frac{4}{49}e^{-7x}+c=-\frac{4x}{7}e^{-7x}+\frac{45}{49}e^{-7x}+c$.

Answer:

$-\frac{4x}{7}e^{-7x}+\frac{45}{49}e^{-7x}+c$