\(\int \cdot \operatorname{se}^{x} d x\)
This is integration by part. We recall the relationship \(\int u d-u x-\int v d u\)
Let \(u=x \cdot \frac{d u}{d x}=1 \Rightarrow d u=d x\)
\begin{aligned}
&d v=e^{x} d x \\
&r=\int e^{x} d x=e x \\
&\int u d y^{\prime}=x \cdot e^{\prime}-\int e^{x} d x \\
&=x e^{x}-e^{x}+c
\end{aligned}
We factorize \(e\)
\begin{aligned}
&e^{x}(x-1)+c \\
&x=\int e^{x} d x=e^{x} \\
&\int w h=x \cdot e^{x}-\int e^{x} d x
\end{aligned}
\(= xe ^{ y }- e ^{ x }+ c\)
We factorize \(e ^{ x }\)
\(e ^{ x }(x-1)+ c\)