Primitive of Root of a x + b

Theorem

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

Proof

Let $u = \sqrt{a x + b}$.

Then:

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

\(=\)

\(\ds \frac 2 a \int u^2 \rd u\)

Primitive of Function of $\sqrt {a x + b}$

\(\ds \)

\(=\)

\(\ds \frac 2 a \frac {u^3} 3\)

Primitive of Power

\(\ds \)

\(=\)

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

substituting for $u$

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.89$
• 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals