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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

$\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {6 a x - 5 b} {24 a^2} \left({\sqrt {a x^2 + b x + c} }\right)^3 + \frac {5 b^2 - 4 a c} {16 a^2} \int \sqrt {a x^2 + b x + c} \ \mathrm d x$


Proof

\(\displaystyle \) \(\) \(\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {2 a x^2} {2 a} \sqrt {a x^2 + b x + c} \ \mathrm d x\) multiplying top and bottom by $2 a$
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {x \left({2 a x + b - b}\right)} {2 a} \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a} \int x \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x - \frac b {2 a} \int x \sqrt {a x^2 + b x + c} \ \mathrm d x\) Linear Combination of Integrals


Let:

\(\displaystyle z\) \(=\) \(\displaystyle a x^2 + b x + c\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle 2 a x + b\) Derivative of Power
\(\displaystyle \implies \ \ \) \(\displaystyle \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x\) \(=\) \(\displaystyle \int \sqrt z \ \mathrm d z\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \left({\sqrt z}\right)^3} 3\) Primitive of Power
\((2):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3\) substituting for $z$


With a view to expressing the primitive $\displaystyle \int x \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x$ in the form:

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

let:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle \int \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3\) from $(2)$ above


Then:

\(\displaystyle \) \(\) \(\displaystyle \int x \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle x \left({\frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3}\right) - \int \left({\frac {2 \left({\sqrt {a x^2 + b x + c} }\right)^3} 3}\right) \times 1 \ \mathrm d x + C\) Integration by Parts
\((3):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \frac {2 x \left({\sqrt {a x^2 + b x + c} }\right)^3} 3 - \frac 2 3 \int \left({\sqrt {a x^2 + b x + c} }\right)^3 \ \mathrm d x + C\) simplifying


Now consider:

\(\displaystyle \) \(\) \(\displaystyle \int \left({\sqrt {a x^2 + b x + c} }\right)^3 \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \left({a x^2 + b x + c}\right) \sqrt {a x^2 + b x + c} \ \mathrm d x\) factorising
\((4):\quad\) \(\displaystyle \) \(=\) \(\displaystyle a \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x + b \int x \sqrt {a x^2 + b x + c} \ \mathrm d x + c \int \sqrt {a x^2 + b x + c} \ \mathrm d x\) Linear Combination of Integrals


Thus:

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


Hence:

\(\displaystyle \) \(\) \(\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x + \frac 1 3 \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\frac {6 a x} {18 a^2} - \frac {3 b} {18 a^2} - \frac {2 b} {18 a^2} }\right) \left({\sqrt {a x^2 + b x + c} }\right)^3\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left({\frac {3 b^2} {12 a^2} - \frac {4 a c} {12 a^2} + \frac {2 b^2} {12 a^2} }\right) \int \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac 4 3 \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x\) \(=\) \(\displaystyle \frac {6 a x - 5 b} {18 a^2} \left({\sqrt {a x^2 + b x + c} }\right)^3 + \frac {5 b^2 - 4 a c} {12 a^2} \int \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x\) \(=\) \(\displaystyle \frac {6 a x - 5 b} {24 a^2} \left({\sqrt {a x^2 + b x + c} }\right)^3 + \frac {5 b^2 - 4 a c} {16 a^2} \int \sqrt {a x^2 + b x + c} \ \mathrm d x\) multiplying by $\dfrac 3 4$

$\blacksquare$

Sources