Integral of $$$5 x e^{- \frac{6 x}{5}}$$$

Find $$$\int 5 x e^{- \frac{6 x}{5}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=5$$$ and $$$f{\left(x \right)} = x e^{- \frac{6 x}{5}}$$$:

$${\color{red}{\int{5 x e^{- \frac{6 x}{5}} d x}}} = {\color{red}{\left(5 \int{x e^{- \frac{6 x}{5}} d x}\right)}}$$

For the integral $$$\int{x e^{- \frac{6 x}{5}} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{- \frac{6 x}{5}} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- \frac{6 x}{5}} d x}=- \frac{5 e^{- \frac{6 x}{5}}}{6}$$$ (steps can be seen »).

Thus,

$$5 {\color{red}{\int{x e^{- \frac{6 x}{5}} d x}}}=5 {\color{red}{\left(x \cdot \left(- \frac{5 e^{- \frac{6 x}{5}}}{6}\right)-\int{\left(- \frac{5 e^{- \frac{6 x}{5}}}{6}\right) \cdot 1 d x}\right)}}=5 {\color{red}{\left(- \frac{5 x e^{- \frac{6 x}{5}}}{6} - \int{\left(- \frac{5 e^{- \frac{6 x}{5}}}{6}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{5}{6}$$$ and $$$f{\left(x \right)} = e^{- \frac{6 x}{5}}$$$:

$$- \frac{25 x e^{- \frac{6 x}{5}}}{6} - 5 {\color{red}{\int{\left(- \frac{5 e^{- \frac{6 x}{5}}}{6}\right)d x}}} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} - 5 {\color{red}{\left(- \frac{5 \int{e^{- \frac{6 x}{5}} d x}}{6}\right)}}$$

Let $$$u=- \frac{6 x}{5}$$$.

Then $$$du=\left(- \frac{6 x}{5}\right)^{\prime }dx = - \frac{6 dx}{5}$$$ (steps can be seen »), and we have that $$$dx = - \frac{5 du}{6}$$$.

So,

$$- \frac{25 x e^{- \frac{6 x}{5}}}{6} + \frac{25 {\color{red}{\int{e^{- \frac{6 x}{5}} d x}}}}{6} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} + \frac{25 {\color{red}{\int{\left(- \frac{5 e^{u}}{6}\right)d u}}}}{6}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{5}{6}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$- \frac{25 x e^{- \frac{6 x}{5}}}{6} + \frac{25 {\color{red}{\int{\left(- \frac{5 e^{u}}{6}\right)d u}}}}{6} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} + \frac{25 {\color{red}{\left(- \frac{5 \int{e^{u} d u}}{6}\right)}}}{6}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$- \frac{25 x e^{- \frac{6 x}{5}}}{6} - \frac{125 {\color{red}{\int{e^{u} d u}}}}{36} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} - \frac{125 {\color{red}{e^{u}}}}{36}$$

Recall that $$$u=- \frac{6 x}{5}$$$:

$$- \frac{25 x e^{- \frac{6 x}{5}}}{6} - \frac{125 e^{{\color{red}{u}}}}{36} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} - \frac{125 e^{{\color{red}{\left(- \frac{6 x}{5}\right)}}}}{36}$$

Therefore,

$$\int{5 x e^{- \frac{6 x}{5}} d x} = - \frac{25 x e^{- \frac{6 x}{5}}}{6} - \frac{125 e^{- \frac{6 x}{5}}}{36}$$

Simplify:

$$\int{5 x e^{- \frac{6 x}{5}} d x} = \frac{25 \left(- 6 x - 5\right) e^{- \frac{6 x}{5}}}{36}$$

Add the constant of integration:

$$\int{5 x e^{- \frac{6 x}{5}} d x} = \frac{25 \left(- 6 x - 5\right) e^{- \frac{6 x}{5}}}{36}+C$$

$$$\int 5 x e^{- \frac{6 x}{5}}\, dx = \frac{25 \left(- 6 x - 5\right) e^{- \frac{6 x}{5}}}{36} + C$$$A