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

Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$

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

Then:

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

Proof

\(\ds \)

\(\)

\(\ds \int x \sqrt {a x^2 + b x + c} \rd x\)

\(\ds \)

\(=\)

\(\ds \int \frac {2 a x \sqrt {a x^2 + b x + c} \rd x} {2 a}\)

multiplying top and bottom by $2 a$

\(\ds \)

\(=\)

\(\ds \int \frac {\paren {2 a x + b - b} \sqrt {a x^2 + b x + c} \rd x} {2 a}\)

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 a} \int \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x\)

Linear Combination of Primitives

Let:

\(\ds z\)

\(=\)

\(\ds a x^2 + b x + c\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d z} {\d x}\)

\(=\)

\(\ds 2 a x + b\)

Derivative of Power

\(\ds \leadsto \ \ \)

\(\ds \int \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x\)

\(=\)

\(\ds \int \sqrt z \rd z\)

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 $z$

So:

\(\ds \)

\(\)

\(\ds \int x \sqrt {a x^2 + b x + c} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac 1 {2 a} \int \paren {2 a x + b} \sqrt {a x^2 + b x + c} \rd x - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 a} \paren {\frac {2 \paren {\sqrt {a x^2 + b x + c} }^3} 3} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x\)

from $(2)$

\(\ds \)

\(=\)

\(\ds \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x\)

simplifying

$\blacksquare$