Primitive of x squared by Root of a x squared plus b x plus c

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Theorem

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

Then:

$\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \rd x = \frac {6 a x - 5 b} {24 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {5 b^2 - 4 a c} {16 a^2} \int \sqrt {a x^2 + b x + c} \rd x$


Proof

\(\ds \) \(\) \(\ds \int x^2 \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {2 a x^2} {2 a} \sqrt {a x^2 + b x + c} \rd x\) multiplying top and bottom by $2 a$
\(\ds \) \(=\) \(\ds \int \frac {x \paren {2 a x + b - b} } {2 a} \sqrt {a x^2 + b x + c} \rd x\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 {2 a} \int x \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x - \frac b {2 a} \int x \sqrt {a x^2 + b x + c} \rd x\) Linear Combination of Integrals


Let:

\(\ds z\) \(=\) \(\ds a x^2 + b x + c\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\rd x}\) \(=\) \(\ds 2 a x + b\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds \int \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x\) \(=\) \(\ds \int \sqrt z \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {2 \paren {\sqrt z}^3} 3\) Primitive of Power
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \frac {2 \paren {\sqrt {a x^2 + b x + c} }^3} 3\) substituting for $z$


With a view to expressing the primitive $\displaystyle \int x \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x$ in the form:

$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 1\) Derivative of Identity Function


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \int \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {2 \paren {\sqrt {a x^2 + b x + c} }^3} 3\) from $(2)$ above


Then:

\(\ds \) \(\) \(\ds \int x \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds x \paren {\frac {2 \paren {\sqrt {a x^2 + b x + c} }^3} 3} - \int \paren {\frac {2 \paren {\sqrt {a x^2 + b x + c} }^3} 3} \times 1 \rd x + C\) Integration by Parts
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \frac {2 x \paren {\sqrt {a x^2 + b x + c} }^3} 3 - \frac 2 3 \int \paren {\sqrt {a x^2 + b x + c} }^3 \rd x + C\) simplifying


Now consider:

\(\ds \) \(\) \(\ds \int \paren {\sqrt {a x^2 + b x + c} }^3 \rd x\)
\(\ds \) \(=\) \(\ds \int \paren {a x^2 + b x + c} \sqrt {a x^2 + b x + c} \rd x\) factorising
\(\text {(4)}: \quad\) \(\ds \) \(=\) \(\ds a \int x^2 \sqrt {a x^2 + b x + c} \rd x + b \int x \sqrt {a x^2 + b x + c} \rd x + c \int \sqrt {a x^2 + b x + c} \rd x\) Linear Combination of Integrals


Thus:

\(\ds \) \(\) \(\ds \int x^2 \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int x \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x - \frac b {2 a} \int x \sqrt {a x^2 + b x + c} \rd x\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\frac {2 x \paren {\sqrt {a x^2 + b x + c} }^3} 3 - \frac 2 3 \int \paren {\sqrt {a x^2 + b x + c} }^3 \rd x}\) from $(3)$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b {2 a} \paren {\frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x}\) Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac 1 {3 a} \int \paren {\sqrt {a x^2 + b x + c} }^3 \rd x\) simplifying
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b {6 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {b^2} {4 a^2} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {6 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {b^2} {4 a^2} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 {3 a} \paren {a \int x^2 \sqrt {a x^2 + b x + c} \rd x + b \int x \sqrt {a x^2 + b x + c} \rd x + c \int \sqrt {a x^2 + b x + c} \rd x}\) from $(4)$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {6 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {b^2} {4 a^2} \int \sqrt {a x^2 + b x + c} \rd x\) simplifying
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x - \frac b {3 a} \int x \sqrt {a x^2 + b x + c} \rd x - \frac c {3 a} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {6 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {b^2} {4 a^2} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x - \frac c {3 a} \int \sqrt {a x^2 + b x + c} \ rd x\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac b {3 a} \paren {\frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x}\) Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$
\(\ds \) \(=\) \(\ds \frac {x \paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {6 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {b^2} {4 a^2} \int \sqrt {a x^2 + b x + c} \rd x\) simplifying
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x - \frac c {3 a} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac {b \paren {\sqrt {a x^2 + b x + c} }^3} {9 a^2} + \frac {b^2} {6 a^2} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \paren {\frac x {3 a} - \frac b {6 a^2} - \frac b {9 a^2} } \paren {\sqrt {a x^2 + b x + c} }^3\) gathering terms
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {\frac {b^2} {4 a^2} - \frac c {3 a} + \frac {b^2} {6 a^2} } \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x\)


Hence:

\(\ds \) \(\) \(\ds \int x^2 \sqrt {a x^2 + b x + c} \rd x + \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \) \(=\) \(\ds \paren {\frac {6 a x} {18 a^2} - \frac {3 b} {18 a^2} - \frac {2 b} {18 a^2} } \paren {\sqrt {a x^2 + b x + c} }^3\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {\frac {3 b^2} {12 a^2} - \frac {4 a c} {12 a^2} + \frac {2 b^2} {12 a^2} } \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \leadsto \ \ \) \(\ds \frac 4 3 \int x^2 \sqrt {a x^2 + b x + c} \rd x\) \(=\) \(\ds \frac {6 a x - 5 b} {18 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {5 b^2 - 4 a c} {12 a^2} \int \sqrt {a x^2 + b x + c} \rd x\)
\(\ds \leadsto \ \ \) \(\ds \int x^2 \sqrt {a x^2 + b x + c} \rd x\) \(=\) \(\ds \frac {6 a x - 5 b} {24 a^2} \paren {\sqrt {a x^2 + b x + c} }^3 + \frac {5 b^2 - 4 a c} {16 a^2} \int \sqrt {a x^2 + b x + c} \rd x\) multiplying by $\dfrac 3 4$

$\blacksquare$


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