Integral of $$$\frac{5}{\left(t^{2} - 25\right)^{2}}$$$

Find $$$\int \frac{5}{\left(t^{2} - 25\right)^{2}}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=5$$$ and $$$f{\left(t \right)} = \frac{1}{\left(t^{2} - 25\right)^{2}}$$$:

$${\color{red}{\int{\frac{5}{\left(t^{2} - 25\right)^{2}} d t}}} = {\color{red}{\left(5 \int{\frac{1}{\left(t^{2} - 25\right)^{2}} d t}\right)}}$$

Perform partial fraction decomposition (steps can be seen »):

$$5 {\color{red}{\int{\frac{1}{\left(t^{2} - 25\right)^{2}} d t}}} = 5 {\color{red}{\int{\left(\frac{1}{500 \left(t + 5\right)} + \frac{1}{100 \left(t + 5\right)^{2}} - \frac{1}{500 \left(t - 5\right)} + \frac{1}{100 \left(t - 5\right)^{2}}\right)d t}}}$$

Integrate term by term:

$$5 {\color{red}{\int{\left(\frac{1}{500 \left(t + 5\right)} + \frac{1}{100 \left(t + 5\right)^{2}} - \frac{1}{500 \left(t - 5\right)} + \frac{1}{100 \left(t - 5\right)^{2}}\right)d t}}} = 5 {\color{red}{\left(\int{\frac{1}{100 \left(t - 5\right)^{2}} d t} - \int{\frac{1}{500 \left(t - 5\right)} d t} + \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + \int{\frac{1}{500 \left(t + 5\right)} d t}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{500}$$$ and $$$f{\left(t \right)} = \frac{1}{t - 5}$$$:

$$5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - 5 {\color{red}{\int{\frac{1}{500 \left(t - 5\right)} d t}}} = 5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - 5 {\color{red}{\left(\frac{\int{\frac{1}{t - 5} d t}}{500}\right)}}$$

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

Then $$$du=\left(t - 5\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

The integral can be rewritten as

$$5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\int{\frac{1}{t - 5} d t}}}}{100} = 5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{100}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{100} = 5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{100}$$

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

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} = - \frac{\ln{\left(\left|{{\color{red}{\left(t - 5\right)}}}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t - 5\right)^{2}} d t} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{100}$$$ and $$$f{\left(t \right)} = \frac{1}{\left(t - 5\right)^{2}}$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + 5 {\color{red}{\int{\frac{1}{100 \left(t - 5\right)^{2}} d t}}} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + 5 {\color{red}{\left(\frac{\int{\frac{1}{\left(t - 5\right)^{2}} d t}}{100}\right)}}$$

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

Then $$$du=\left(t - 5\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

So,

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{\left(t - 5\right)^{2}} d t}}}}{20} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{20}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{20}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{u^{-2} d u}}}}{20}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{20}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{20}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{20}$$

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

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{u}}^{-1}}{20} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{100 \left(t + 5\right)^{2}} d t} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\left(t - 5\right)}}^{-1}}{20}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{100}$$$ and $$$f{\left(t \right)} = \frac{1}{\left(t + 5\right)^{2}}$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + 5 {\color{red}{\int{\frac{1}{100 \left(t + 5\right)^{2}} d t}}} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + 5 {\color{red}{\left(\frac{\int{\frac{1}{\left(t + 5\right)^{2}} d t}}{100}\right)}} - \frac{1}{20 \left(t - 5\right)}$$

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

Then $$$du=\left(t + 5\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

So,

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{\left(t + 5\right)^{2}} d t}}}}{20} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{20} - \frac{1}{20 \left(t - 5\right)}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{20} - \frac{1}{20 \left(t - 5\right)}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\int{u^{-2} d u}}}}{20} - \frac{1}{20 \left(t - 5\right)}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{20} - \frac{1}{20 \left(t - 5\right)}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{20} - \frac{1}{20 \left(t - 5\right)}=- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{20} - \frac{1}{20 \left(t - 5\right)}$$

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

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{u}}^{-1}}{20} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 \int{\frac{1}{500 \left(t + 5\right)} d t} - \frac{{\color{red}{\left(t + 5\right)}}^{-1}}{20} - \frac{1}{20 \left(t - 5\right)}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{500}$$$ and $$$f{\left(t \right)} = \frac{1}{t + 5}$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 {\color{red}{\int{\frac{1}{500 \left(t + 5\right)} d t}}} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + 5 {\color{red}{\left(\frac{\int{\frac{1}{t + 5} d t}}{500}\right)}} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)}$$

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

Then $$$du=\left(t + 5\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

The integral becomes

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{{\color{red}{\int{\frac{1}{t + 5} d t}}}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)}$$

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

$$- \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{\ln{\left(\left|{{\color{red}{\left(t + 5\right)}}}\right| \right)}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)}$$

Therefore,

$$\int{\frac{5}{\left(t^{2} - 25\right)^{2}} d t} = - \frac{\ln{\left(\left|{t - 5}\right| \right)}}{100} + \frac{\ln{\left(\left|{t + 5}\right| \right)}}{100} - \frac{1}{20 \left(t + 5\right)} - \frac{1}{20 \left(t - 5\right)}$$

Simplify:

$$\int{\frac{5}{\left(t^{2} - 25\right)^{2}} d t} = \frac{- 10 t + \left(t - 5\right) \left(t + 5\right) \left(- \ln{\left(\left|{t - 5}\right| \right)} + \ln{\left(\left|{t + 5}\right| \right)}\right)}{100 \left(t - 5\right) \left(t + 5\right)}$$

Add the constant of integration:

$$\int{\frac{5}{\left(t^{2} - 25\right)^{2}} d t} = \frac{- 10 t + \left(t - 5\right) \left(t + 5\right) \left(- \ln{\left(\left|{t - 5}\right| \right)} + \ln{\left(\left|{t + 5}\right| \right)}\right)}{100 \left(t - 5\right) \left(t + 5\right)}+C$$

$$$\int \frac{5}{\left(t^{2} - 25\right)^{2}}\, dt = \frac{- 10 t + \left(t - 5\right) \left(t + 5\right) \left(- \ln\left(\left|{t - 5}\right|\right) + \ln\left(\left|{t + 5}\right|\right)\right)}{100 \left(t - 5\right) \left(t + 5\right)} + C$$$A