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

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Theorem

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

Then:

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


Proof

\(\ds \int \frac {x^2 \rd x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac 1 a \paren {1 - \frac {b x + c} {a x^2 + b x + c} } \rd x\) by division
\(\ds \) \(=\) \(\ds \frac 1 a \int \rd x - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac x a - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac x a - \frac b a \paren {\frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {a x^2 + b x + c}\) Primitive of $\dfrac x {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} + \frac {b^2} {2 a^2} \int \frac {\d x} {a x^2 + b x + c} - \frac {2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}\) multiplying out
\(\ds \) \(=\) \(\ds \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} + \frac {b^2 - 2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}\) simplification

$\blacksquare$


Examples

Primitive of $\dfrac {6 x^2 + 10 x + 5} {3 x^2 + 4 x + 2}$

$\ds \int \dfrac {6 x^2 + 10 x + 5} {3 x^2 + 4 x + 2} \rd x = 2 x + \dfrac 1 3 \ln \size {3 x^3 + 4 x + 2} - \dfrac 1 {3 \sqrt 2} \arctan \dfrac {3 x + 2} {\sqrt 2} + C$


Also see


Sources

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