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

## 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

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

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

$\blacksquare$