Primitive of x by Power of a x + b

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Theorem

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

where $n \ne - 1$ and $n \ne - 2$.


Proof

Let $u = a x + b$.

Then:

\(\ds x\) \(=\) \(\ds \frac {u - b} a\)
\(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 a\)


Then:

\(\ds \int x \paren {a x + b}^n \rd x\) \(=\) \(\ds \frac 1 a \int \frac {u - b} a u^n \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \int u^{n + 1} \rd u - \frac b {a^2} \int u^n \rd u\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \frac {u^{n + 2} } {n + 2} - \frac b {a^2} \frac {u^{n + 1} } {n + 1} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^{n + 2} } {\paren {n + 2} a^2} - \frac {b \paren {a x + b}^{n + 1} } {\paren {n + 1} a^2} + C\) substituting for $u$

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Also see


Sources

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