Integral of $$$\frac{1}{x^{2} \left(a^{2} + x^{2}\right)}$$$ with respect to $$$x$$$
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Find $$$\int \frac{1}{x^{2} \left(a^{2} + x^{2}\right)}\, dx$$$.
Solution
Perform partial fraction decomposition:
$${\color{red}{\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x}}} = {\color{red}{\int{\left(- \frac{1}{a^{2} \left(a^{2} + x^{2}\right)} + \frac{1}{a^{2} x^{2}}\right)d x}}}$$
Integrate term by term:
$${\color{red}{\int{\left(- \frac{1}{a^{2} \left(a^{2} + x^{2}\right)} + \frac{1}{a^{2} x^{2}}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{a^{2} x^{2}} d x} - \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{a^{2}}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:
$$- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + {\color{red}{\int{\frac{1}{a^{2} x^{2}} d x}}} = - \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + {\color{red}{\frac{\int{\frac{1}{x^{2}} d x}}{a^{2}}}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:
$$- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x^{2}} d x}}}}{a^{2}}=- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + \frac{{\color{red}{\int{x^{-2} d x}}}}{a^{2}}=- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + \frac{{\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{a^{2}}=- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + \frac{{\color{red}{\left(- x^{-1}\right)}}}{a^{2}}=- \int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x} + \frac{{\color{red}{\left(- \frac{1}{x}\right)}}}{a^{2}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{a^{2}}$$$ and $$$f{\left(x \right)} = \frac{1}{a^{2} + x^{2}}$$$:
$$- {\color{red}{\int{\frac{1}{a^{2} \left(a^{2} + x^{2}\right)} d x}}} - \frac{1}{a^{2} x} = - {\color{red}{\frac{\int{\frac{1}{a^{2} + x^{2}} d x}}{a^{2}}}} - \frac{1}{a^{2} x}$$
Let $$$u=\frac{x}{\left|{a}\right|}$$$.
Then $$$du=\left(\frac{x}{\left|{a}\right|}\right)^{\prime }dx = \frac{dx}{\left|{a}\right|}$$$ (steps can be seen »), and we have that $$$dx = \left|{a}\right| du$$$.
The integral can be rewritten as
$$- \frac{{\color{red}{\int{\frac{1}{a^{2} + x^{2}} d x}}}}{a^{2}} - \frac{1}{a^{2} x} = - \frac{{\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(u^{2} + 1\right)} d u}}}}{a^{2}} - \frac{1}{a^{2} x}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\left|{a}\right|}{a^{2}}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$:
$$- \frac{{\color{red}{\int{\frac{\left|{a}\right|}{a^{2} \left(u^{2} + 1\right)} d u}}}}{a^{2}} - \frac{1}{a^{2} x} = - \frac{{\color{red}{\frac{\left|{a}\right| \int{\frac{1}{u^{2} + 1} d u}}{a^{2}}}}}{a^{2}} - \frac{1}{a^{2} x}$$
The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:
$$- \frac{1}{a^{2} x} - \frac{\left|{a}\right| {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{a^{4}} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| {\color{red}{\operatorname{atan}{\left(u \right)}}}}{a^{4}}$$
Recall that $$$u=\frac{x}{\left|{a}\right|}$$$:
$$- \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{u}} \right)}}{a^{4}} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left({\color{red}{\frac{x}{\left|{a}\right|}}} \right)}}{a^{4}}$$
Therefore,
$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = - \frac{1}{a^{2} x} - \frac{\left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4}}$$
Simplify:
$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x}$$
Add the constant of integration:
$$\int{\frac{1}{x^{2} \left(a^{2} + x^{2}\right)} d x} = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x}+C$$
Answer
$$$\int \frac{1}{x^{2} \left(a^{2} + x^{2}\right)}\, dx = \frac{- a^{2} - x \left|{a}\right| \operatorname{atan}{\left(\frac{x}{\left|{a}\right|} \right)}}{a^{4} x} + C$$$A