Primitive of x by Half Integer Power of a x squared plus b x plus c

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Theorem

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

Then:

$\ds \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x = \frac {\paren {a x^2 + b x + c}^{n + \frac 3 2} } {a \paren {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x$


Proof

\(\ds \) \(\) \(\ds \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {2 a x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x} {2 a}\) multiplying top and bottom by $2 a$
\(\ds \) \(=\) \(\ds \int \frac {\paren {2 a x + b - b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x} {2 a}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) 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
\(\ds \leadsto \ \ \) \(\ds \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) \(=\) \(\ds \int z^{n + \frac 1 2} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac {z^{n + \frac 3 2} } {n + \frac 3 2}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {2 z^{n + \frac 3 2} } {2 n + 3}\) simplifying
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \frac {2 \paren {a x^2 + b x + c}^{n + \frac 3 2} } {2 n + 3}\) substituting for $z$


So:

\(\ds \) \(\) \(\ds \int x \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \paren {2 a x + b} \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) from $(1)$
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \paren {\frac {2 \paren {a x^2 + b x + c}^{n + \frac 3 2} } {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) from $(2)$
\(\ds \) \(=\) \(\ds \frac {\paren {a x^2 + b x + c}^{n + \frac 3 2} } {a \paren {2 n + 3} } - \frac b {2 a} \int \paren {a x^2 + b x + c}^{n + \frac 1 2} \rd x\) simplifying

$\blacksquare$


Sources

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