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

## Theorem

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

Then:

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

## Proof

 $\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}$ $=$ $\ds \int \frac {c \rd x} {c x \paren {a x^2 + b x + c}^2}$ multiplying top and bottom by $c$ $\ds$ $=$ $\ds \frac 1 c \int \frac {c \rd x} {x \paren {a x^2 + b x + c}^2}$ Primitive of Constant Multiple of Function $\ds$ $=$ $\ds \frac 1 c \int \frac {a x^2 + b x + c - a x^2 - b x} {x \paren {a x^2 + b x + c}^2} \rd x$ adding and subtracting $a x^2 + b x$ $\ds$ $=$ $\ds \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x \paren {a x^2 + b x + c}^2}$ Linear Combination of Integrals $\ds$  $\, \ds - \,$ $\ds \frac b c \int \frac {x \rd x} {x \paren {a x^2 + b x + c}^2}$ $\ds$ $=$ $\ds \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2}$ simplification $\ds$  $\, \ds - \,$ $\ds \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}$ $\ds$ $=$ $\ds - \frac {2 a} {2 c} \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}$ splitting up the $\dfrac b c$ term $\ds$  $\, \ds - \,$ $\ds \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$ $\ds$ $=$ $\ds - \frac 1 {2 c} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^2}$ Linear Combination of Integrals $\ds$  $\, \ds - \,$ $\ds \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$ $\ds$ $=$ $\ds -\frac 1 {2 c} \paren {\frac {-1} {a x^2 + b x + c} }$ Primitive of Function under its Derivative $\ds$  $\, \ds - \,$ $\ds \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$ $\ds$ $=$ $\ds \frac 1 {2 c \paren {a x^2 + b x + c} }$ simplifying $\ds$  $\, \ds - \,$ $\ds \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$

$\blacksquare$