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

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Theorem

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

Then:

$\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^2} = \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$


Proof

Let:

\(\ds z\) \(=\) \(\ds 2 a x + b\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a\) Derivative of Power


Then:

\(\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) \(=\) \(\ds \int \paren {\frac {4 a} {\paren {2 a x + b}^2 + 4 a c - b^2} }^2 \rd x\) Completing the Square
\(\ds \) \(=\) \(\ds \int \frac {\paren {4 a}^2} {\paren {z^2 + 4 a c - b^2}^2} \frac {\d z} {2 a}\) Integration by Substitution
\(\ds \) \(=\) \(\ds 8 a \int \frac {\d z} {\paren {z^2 + 4 a c - b^2}^2}\) simplifying


Let $u = z^2$.

Let:

\(\ds u\) \(=\) \(\ds z^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d z}\) \(=\) \(\ds 2 z\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds 8 a \int \frac {\d z} {\paren {z^2 + 4 a c - b^2}^2}\) \(=\) \(\ds 8 a \int \frac {\d u} {2 \sqrt u \paren {u + 4 a c - b^2}^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds 4 a \int \frac {\d u} {\sqrt u \paren {u + 4 a c - b^2}^2}\) simplifying

Recall the result Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:

$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$


Let:

\(\ds x\) \(=:\) \(\ds u\)
\(\ds a\) \(=:\) \(\ds 1\)
\(\ds b\) \(=:\) \(\ds 0\)
\(\ds p\) \(=:\) \(\ds 1\)
\(\ds q\) \(=:\) \(\ds 4 a c - b^2\)
\(\ds n\) \(=:\) \(\ds 2\)


Then:

\(\ds 4 a \int \frac {\d u} {\sqrt u \paren {u + 4 a c - b^2}^2}\) \(=\) \(\ds 4 a \paren {\frac {\sqrt u} {\paren {4 a c - b^2} \paren {u + 4 a c - b^2} } + \frac 1 {2 \paren {4 a c - b^2} } \int \frac {\d u} {\paren {u + 4 a c - b^2} \sqrt u} }\)
\(\ds \) \(=\) \(\ds 4 a \paren {\frac z {\paren {4 a c - b^2} \paren {z^2 + 4 a c - b^2} } + \frac 1 {2 \paren {4 a c - b^2} } \int \frac {2 z \rd z} {\paren {z^2 + 4 a c - b^2} z} }\) substituting for $z$
\(\ds \) \(=\) \(\ds 4 a \frac {2 a x + b} {\paren {4 a c - b^2} \paren {\paren {2 a x + b}^2 + 4 a c - b^2} } + \frac {2 a} {4 a c - b^2} \int \frac {4 a \rd x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2} }\) substituting for $x$
\(\ds \) \(=\) \(\ds \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {\paren {a x^2 + b x + c} }\) Completing the Square

$\blacksquare$


Sources

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