Integral of $$$\frac{1}{- a^{2} + x^{2}}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{- a^{2} + x^{2}}\, dx$$$.

Perform partial fraction decomposition:

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

Integrate term by term:

$${\color{red}{\int{\left(- \frac{1}{2 \left(x + \left|{a}\right|\right) \left|{a}\right|} + \frac{1}{2 \left(x - \left|{a}\right|\right) \left|{a}\right|}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{2 \left(x - \left|{a}\right|\right) \left|{a}\right|} d x} - \int{\frac{1}{2 \left(x + \left|{a}\right|\right) \left|{a}\right|} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2 \left|{a}\right|}$$$ and $$$f{\left(x \right)} = \frac{1}{- a + x}$$$:

$$- \int{\frac{1}{2 \left(x + \left|{a}\right|\right) \left|{a}\right|} d x} + {\color{red}{\int{\frac{1}{2 \left(x - \left|{a}\right|\right) \left|{a}\right|} d x}}} = - \int{\frac{1}{2 \left(x + \left|{a}\right|\right) \left|{a}\right|} d x} + {\color{red}{\left(\frac{\int{\frac{1}{- a + x} d x}}{2 \left|{a}\right|}\right)}}$$

Let $$$u=- a + x$$$.

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

The integral becomes

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

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

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

Recall that $$$u=- a + x$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2 \left|{a}\right|}$$$ and $$$f{\left(x \right)} = \frac{1}{a + x}$$$:

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

Let $$$u=a + x$$$.

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

Thus,

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

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

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

Recall that $$$u=a + x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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