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

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Theorem

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

Then:

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


Proof

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

$\blacksquare$


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