## Problem Statement

Calculate the indefinite integral:

$$
\int(1-6 x) e^{2 x} d x
$$
## Solution

$$
\int(1-6 x) e^{2 x} d x=
$$

Let's denote:

$$
\begin{aligned}
& u=1-6 x ; d u=-6 d x \\
& d v=e^{2 x} d x ; v=\frac{1}{2} e^{2 x}
\end{aligned}
$$

Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:

$$
\begin{aligned}
& =(1-6 x) \cdot \frac{1}{2} e^{2 x}-\int \frac{1}{2} e^{2 x} \cdot(-6) d x=(1-6 x) \cdot \frac{1}{2} e^{2 x}+\int 3 e^{2 x} d x= \\
& =(1-6 x) \cdot \frac{1}{2} e^{2 x}+\frac{3}{2} \cdot e^{2 x}+C=(2-3 x) e^{2 x}+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-7$ >

Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

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## Problem Kuznetsov Integrals 1-8

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