Integral of $$$t^{2} e^{5 t}$$$

Find $$$\int t^{2} e^{5 t}\, dt$$$.

For the integral $$$\int{t^{2} e^{5 t} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=t^{2}$$$ and $$$\operatorname{dv}=e^{5 t} dt$$$.

Then $$$\operatorname{du}=\left(t^{2}\right)^{\prime }dt=2 t dt$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{5 t} d t}=\frac{e^{5 t}}{5}$$$ (steps can be seen »).

The integral becomes

$${\color{red}{\int{t^{2} e^{5 t} d t}}}={\color{red}{\left(t^{2} \cdot \frac{e^{5 t}}{5}-\int{\frac{e^{5 t}}{5} \cdot 2 t d t}\right)}}={\color{red}{\left(\frac{t^{2} e^{5 t}}{5} - \int{\frac{2 t e^{5 t}}{5} d t}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{2}{5}$$$ and $$$f{\left(t \right)} = t e^{5 t}$$$:

$$\frac{t^{2} e^{5 t}}{5} - {\color{red}{\int{\frac{2 t e^{5 t}}{5} d t}}} = \frac{t^{2} e^{5 t}}{5} - {\color{red}{\left(\frac{2 \int{t e^{5 t} d t}}{5}\right)}}$$

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

Let $$$\operatorname{u}=t$$$ and $$$\operatorname{dv}=e^{5 t} dt$$$.

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

The integral can be rewritten as

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 {\color{red}{\int{t e^{5 t} d t}}}}{5}=\frac{t^{2} e^{5 t}}{5} - \frac{2 {\color{red}{\left(t \cdot \frac{e^{5 t}}{5}-\int{\frac{e^{5 t}}{5} \cdot 1 d t}\right)}}}{5}=\frac{t^{2} e^{5 t}}{5} - \frac{2 {\color{red}{\left(\frac{t e^{5 t}}{5} - \int{\frac{e^{5 t}}{5} d t}\right)}}}{5}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(t \right)} = e^{5 t}$$$:

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\int{\frac{e^{5 t}}{5} d t}}}}{5} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\left(\frac{\int{e^{5 t} d t}}{5}\right)}}}{5}$$

Let $$$u=5 t$$$.

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

So,

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\int{e^{5 t} d t}}}}{25} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\int{\frac{e^{u}}{5} d u}}}}{25}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\int{\frac{e^{u}}{5} d u}}}}{25} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\left(\frac{\int{e^{u} d u}}{5}\right)}}}{25}$$

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

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{\int{e^{u} d u}}}}{125} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 {\color{red}{e^{u}}}}{125}$$

Recall that $$$u=5 t$$$:

$$\frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 e^{{\color{red}{u}}}}{125} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 e^{{\color{red}{\left(5 t\right)}}}}{125}$$

Therefore,

$$\int{t^{2} e^{5 t} d t} = \frac{t^{2} e^{5 t}}{5} - \frac{2 t e^{5 t}}{25} + \frac{2 e^{5 t}}{125}$$

Simplify:

$$\int{t^{2} e^{5 t} d t} = \frac{\left(25 t^{2} - 10 t + 2\right) e^{5 t}}{125}$$

Add the constant of integration:

$$\int{t^{2} e^{5 t} d t} = \frac{\left(25 t^{2} - 10 t + 2\right) e^{5 t}}{125}+C$$

$$$\int t^{2} e^{5 t}\, dt = \frac{\left(25 t^{2} - 10 t + 2\right) e^{5 t}}{125} + C$$$A