Primitive of x squared by Power of a x + b

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int x^2 \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 3} } {\paren {n + 3} a^3} - \frac {2 b \paren {a x + b}^{n + 2} } {\paren {n + 2} a^3} + \frac {b^2 \paren {a x + b}^{n + 1} } {\paren {n + 1} a^3} + C$

where $n \notin \set {-1, -2, -3}$.


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:

\(\ds \int x \paren {a x + b}^n \rd x\) \(=\) \(\ds \int \frac 1 a \paren {\frac {u - b} a}^2 u^n \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac 1 {a^3} \paren {u^{n + 2} - 2 b u^{n + 1} + b^2 u^n} \rd u\) Square of Difference and multiplying out
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int u^{n + 3} \rd u - \frac {2 b} {a^3} \int u^{n + 1} \rd u + \frac {b^2} {a^3} \int u^n \rd u\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \frac {u^{n + 3} } {n + 3} - \frac {2 b} {a^3} \frac {u^{n + 2} } {n + 2} + \frac {b^2} {a^3} \frac {u^{n + 1} } {n + 1} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^{n + 3} } {\paren {n + 3} a^3} - \frac {2 b \paren {a x + b}^{n + 2} } {\paren {n + 2} a^3} + \frac {b^2 \paren {a x + b}^{n + 1} } {\paren {n + 1} a^3} + C\) substituting for $u$

$\blacksquare$


Also see


Sources