Primitive of Power of a x + b

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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 1

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 $u$

$\blacksquare$


Proof 2

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 $u$

$\blacksquare$


Proof 3

\(\ds \map {\dfrac \d {\d x} } {\frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} }\) \(=\) \(\ds \dfrac {\paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a} \map {\dfrac \d {\d x} } {a x + b}\) Power Rule for Derivatives, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {a \paren {n + 1} \paren {a x + b}^n} {\paren {n + 1} a}\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds \paren {a x + b}^n\) simplifying

The result follows by definition of primitive.

$\blacksquare$


Also see


Sources

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