Primitive of x squared over a x + b squared/Proof 1
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Theorem
- $\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2} = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C$
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:
\(\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2}\) | \(=\) | \(\ds \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac 1 {a^3} \paren {1 - \frac {2 b} u + \frac {b^2} {u^2} } \rd u\) | Square of Difference, and simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a^3} \int \d u - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac u {a^3} - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C\) | Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C\) | Primitive of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \frac {-1} u + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C\) | substituting for $u$ and rearranging |
$\blacksquare$