Integral of $$$\frac{i n t}{- x^{3} + x}$$$ with respect to $$$x$$$

Find $$$\int \frac{i n t}{- x^{3} + x}\, dx$$$.

Simplify the integrand:

$${\color{red}{\int{\frac{i n t}{- x^{3} + x} d x}}} = {\color{red}{\int{\frac{i n t}{x \left(1 - x^{2}\right)} d x}}}$$

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

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

Let $$$u=1 - x^{2}$$$.

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

So,

$$i n t {\color{red}{\int{\frac{1}{x \left(1 - x^{2}\right)} d x}}} = i n t {\color{red}{\int{\frac{1}{2 u \left(u - 1\right)} d u}}}$$

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

$$i n t {\color{red}{\int{\frac{1}{2 u \left(u - 1\right)} d u}}} = i n t {\color{red}{\left(\frac{\int{\frac{1}{u \left(u - 1\right)} d u}}{2}\right)}}$$

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

$$\frac{i n t {\color{red}{\int{\frac{1}{u \left(u - 1\right)} d u}}}}{2} = \frac{i n t {\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}}{2}$$

Integrate term by term:

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

Let $$$v=u - 1$$$.

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

Thus,

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

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

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

Recall that $$$v=u - 1$$$:

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

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

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

Recall that $$$u=1 - x^{2}$$$:

$$\frac{i n t \left(\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)}\right)}{2} = \frac{i n t \left(\ln{\left(\left|{-1 + {\color{red}{\left(1 - x^{2}\right)}}}\right| \right)} - \ln{\left(\left|{{\color{red}{\left(1 - x^{2}\right)}}}\right| \right)}\right)}{2}$$

Therefore,

$$\int{\frac{i n t}{- x^{3} + x} d x} = \frac{i n t \left(\ln{\left(x^{2} \right)} - \ln{\left(\left|{x^{2} - 1}\right| \right)}\right)}{2}$$

Simplify:

$$\int{\frac{i n t}{- x^{3} + x} d x} = \frac{i n t \left(2 \ln{\left(x \right)} - \ln{\left(\left|{x^{2} - 1}\right| \right)}\right)}{2}$$

Add the constant of integration:

$$\int{\frac{i n t}{- x^{3} + x} d x} = \frac{i n t \left(2 \ln{\left(x \right)} - \ln{\left(\left|{x^{2} - 1}\right| \right)}\right)}{2}+C$$

$$$\int \frac{i n t}{- x^{3} + x}\, dx = \frac{i n t \left(2 \ln\left(x\right) - \ln\left(\left|{x^{2} - 1}\right|\right)\right)}{2} + C$$$A