Primitive of Power of a x + b/Proof 2

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:

$\dfrac {\d u} {\d x} = a$

Then:

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

\(=\)

\(\ds \int \dfrac {u^n} a \rd u\)

Integration by Substitution

\(\ds \)

\(=\)

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

Primitive of Power

\(\ds \)

\(=\)

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

substituting back for $u$

$\blacksquare$

Sources

• 1944: R.P. Gillespie: Integration (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 8$. Change of Variable