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

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Theorem

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

Then:

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


Proof

Let:

\(\ds z\) \(=\) \(\ds a x^2 + b x + c\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a x + b\) Derivative of Power


Then:

\(\ds \) \(\) \(\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {2 a x \rd x} {\paren {a x^2 + b x + c}^2}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\paren {a x^2 + b x + c}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd z} {\paren {2 a x + b} z^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) Integration by Substitution from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\d z} {z^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) simplification
\(\ds \) \(=\) \(\ds \frac {-1} {2 a z} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {-1} {2 a z} - \frac b {2 a} \paren {\frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c} }\) Primitive of $\dfrac 1 {\paren {a x^2 + b x + c}^2}$
\(\ds \) \(=\) \(\ds \frac {-1} {2 a \paren {a x^2 + b x + c} } - \frac {b \paren {2 a x + b} } {2 a \paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {b^2 - 4 a c - 2 a b x - b^2} {2 a \paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}\) common denominator
\(\ds \) \(=\) \(\ds \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}\) simplification

$\blacksquare$


Sources

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