Primitive of Reciprocal of x squared by square of a x squared plus b x plus c

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\displaystyle \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2} = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}$


Proof 1

From Primitive of Reciprocal of Power of x by Power of a x squared plus b x plus c:

\(\displaystyle \int \frac {\d x} {x^m \paren {a x^2 + b x + c}^n}\) \(=\) \(\displaystyle \frac {-1} {\paren {m - 1} c x^{m - 1} \paren {a x^2 + b x + c}^{n - 1} }\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {\paren {m + 2 n - 3} a} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 2} \paren {a x^2 + b x + c}^n}\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {\paren {m - n + 2} b} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 1} \paren {a x^2 + b x + c}^n}\)


Setting $m = n = 2$:

\(\displaystyle \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}\) \(=\) \(\displaystyle \frac {-1} {\paren {2 - 1} c x^{2 - 1} \paren {a x^2 + b x + c}^{2 - 1} }\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {\paren {2 + 2 \times 2 - 3} a} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 2} \paren {a x^2 + b x + c}^2}\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {\paren {2 - 2 + 2} b} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 1} \paren {a x^2 + b x + c}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\) simplifying

$\blacksquare$


Proof 2

First:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {c \rd x} {c x^2 \paren {a x^2 + b x + c}^2}\) multiplying top and bottom by $c$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \int \frac {c \rd x} {x^2 \paren {a x^2 + b x + c}^2}\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \int \frac {a x^2 + b x + c - a x^2 - b x} {x^2 \paren {a x^2 + b x + c}^2} \rd x\) adding and subtracting $a x^2 + b x$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x^2 \paren {a x^2 + b x + c}^2}\) Linear Combination of Integrals
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac b c \int \frac {x \rd x} {x^2 \paren {a x^2 + b x + c}^2}\)
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) simplification
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\)


Next, with a view to obtaining an expression in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \frac 1 {\paren {a x^2 + b x + c} }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a x^2 + b x + c}^{-1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d u} {\d x}\) \(=\) \(\displaystyle -\paren {2 a x + b} \paren {a x^2 + b x + c}^{-2}\) Chain Rule for Derivatives and Derivative of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2}\) simplifying


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac 1 {x^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} x\) Primitive of Power


Then:

\(\displaystyle \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} }\) \(=\) \(\displaystyle \int \frac 1 {\paren {a x^2 + b x + c} } \frac 1 {x^2} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\paren {a x^2 + b x + c} } \frac {-1} x - \int \frac {-1} x \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2} \rd x\) Integration by Parts
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) simplification
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\) Linear Combination of Integrals


Thus:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\) from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 c \paren {\frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} }\) from $(2)$
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}\) gathering terms

$\blacksquare$


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