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

## 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 $\dfrac 3 4$

$\blacksquare$