Primitive of x squared over a x + b/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

Put $u = a x + b$.

Then:

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


Then:

\(\ds \int \frac {x^2 \rd x} {a x + b}\) \(=\) \(\ds \int \frac 1 a \paren {\frac {u - b} a}^2 \frac {\d u} u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \frac {\paren {u - b}^2} u \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \paren {u - 2 b + \frac {b^2} u} \rd u\) multiplying out and simplifying
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\int u \rd u - \int 2 b \rd u + \int \frac {b^2} u \rd u}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\frac {u^2} 2 + C - \int 2 b \rd u + \int \frac {b^2} u \rd u}\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + C + \int \frac {b^2} u \rd u}\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + C + b^2 \int \frac {\d u} u}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + b^2 \ln \size u} + C\) Primitive of Reciprocal and subsuming arbitrary constant $C$
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\) substituting for $u$ and simplifying

$\blacksquare$