## Problem Statement

Calculate the indefinite integral:

$$
\int(4-3 x) e^{-3 x} d x
$$
## Solution

$$
\int(4-3 x) e^{-3 x} d x=
$$

Let's denote:

$$
\begin{aligned}
& u=4-3 x ; d u=-3 d x \\
& d v=e^{-3 x} d x ; v=-\frac{1}{3} e^{-3 x}
\end{aligned}
$$

We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:

$$
\begin{aligned}
& =(4-3 x) \cdot\left(-\frac{1}{3} e^{-3 x}\right)-\int\left(-\frac{1}{3} e^{-3 x}\right) \cdot(-3) d x= \\
& =-\frac{1}{3}(4-3 x) e^{-3 x}-\int e^{-3 x} d x=-\frac{1}{3}(4-3 x) e^{-3 x}-\left(-\frac{1}{3} e^{-3 x}\right)+C= \\
& =-\frac{1}{3}(3-3 x) e^{-3 x}+C=(x-1) e^{-3 x}+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-1$ »

Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 1-2

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