Primitive of Power of a x + b/Proof 1

Theorem

$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$

where $n \ne 1$.

Proof

Let $u = a x + b$.

Then:

\(\ds \int \paren {a x + b}^n \rd x\)

\(=\)

\(\ds \frac 1 a \int u^n \rd u\)

Primitive of Function of $a x + b$

\(\ds \)

\(=\)

\(\ds \frac 1 a \frac {u^{n + 1} } {n + 1} + C\)

Primitive of Power

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C\)

substituting for $u$

$\blacksquare$

Sources

• 1945: A. Geary, H.V. Lowry and H.A. Hayden: Advanced Mathematics for Technical Students, Part I ... (previous) ... (next): Chapter $\text {III}$: Integration: Three rules for integration: $\text {III}$