Integral of $$$\frac{t^{2}}{t^{2} - 2}$$$

Find $$$\int \frac{t^{2}}{t^{2} - 2}\, dt$$$.

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$\int{\frac{2}{t^{2} - 2} d t} + {\color{red}{\int{1 d t}}} = \int{\frac{2}{t^{2} - 2} d t} + {\color{red}{t}}$$

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

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

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

$$t + 2 {\color{red}{\int{\frac{1}{t^{2} - 2} d t}}} = t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}}$$

Integrate term by term:

$$t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}} = t + 2 {\color{red}{\left(\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{\sqrt{2}}{4}$$$ and $$$f{\left(t \right)} = \frac{1}{t + \sqrt{2}}$$$:

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}}} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t + \sqrt{2}} d t}}{4}\right)}}$$

Let $$$u=t + \sqrt{2}$$$.

Then $$$du=\left(t + \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

Thus,

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{t + \sqrt{2}} d t}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recall that $$$u=t + \sqrt{2}$$$:

$$t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} = t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(t + \sqrt{2}\right)}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{\sqrt{2}}{4}$$$ and $$$f{\left(t \right)} = \frac{1}{t - \sqrt{2}}$$$:

$$t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}}} = t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t - \sqrt{2}} d t}}{4}\right)}}$$

Let $$$u=t - \sqrt{2}$$$.

Then $$$du=\left(t - \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (steps can be seen »), and we have that $$$dt = du$$$.

The integral becomes

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

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

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

Recall that $$$u=t - \sqrt{2}$$$:

$$t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(t - \sqrt{2}\right)}}}\right| \right)}}{2}$$

Therefore,

$$\int{\frac{t^{2}}{t^{2} - 2} d t} = t + \frac{\sqrt{2} \ln{\left(\left|{t - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2}$$

Add the constant of integration:

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

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