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:

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

Integrate term by term:

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

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

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

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 can be rewritten as

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

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

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

Recall that $u=a + x$:

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

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

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

Let $u=- a + x$.

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

So,

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

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

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

Recall that $u=- a + x$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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