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

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Theorem

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

Then:

$\ds \int \frac {x \rd x} {a x^2 + b x + c} = \frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}$


Proof

First note that by Derivative of Power:

$(1): \quad \map {\dfrac \d {\d x} } {a x^2 + b x + c} = 2 a x + b$


Then:

\(\ds \int \frac {x \rd x} {a x^2 + b x + c}\) \(=\) \(\ds \frac 1 {2 a} \int \frac {2 a x \rd x} {a x^2 + b x + c}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {a x^2 + b x + c}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}\) Primitive of Function under its Derivative and $(1)$

$\blacksquare$


Examples

Primitive of $\dfrac x {x^2 + 4 x + 5}$

$\ds \int \dfrac {x \rd x} {x^2 + 4 x + 5} = \frac 1 2 \map \ln {x^2 + 4 x + 5} - 2 \map \arctan {x + 2} + C$


Also see


Sources

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