Primitive of Reciprocal of x squared by square of a x squared plus b x plus c/Proof 2

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Theorem

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

Then:

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


Proof

First:

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


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

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

let:

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


and let:

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


Then:

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


Thus:

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

$\blacksquare$