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

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Theorem

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

Then:

$\ds \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } + C$


Proof

First:

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


Then:

\(\ds \int \frac {x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) \(=\) \(\ds \frac 1 {2 a} \int \frac {2 a x \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\d z} {z^{3/2} } - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\frac {-2} {\sqrt z} } - \frac b {2 a} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt z} - \frac b {2 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } } + C\) Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {a x^2 + b x + c} } - \frac b {2 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac {2 \paren {b x + 2 c} } {\paren {b^2 - 4 a c} \sqrt {a x^2 + b x + c} } + C\) simplifying

$\blacksquare$


Sources

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