Primitive of Reciprocal of x squared by a x + b squared/Partial Fraction Expansion
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Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x + b}^2}$
- $\dfrac 1 {x^2 \paren {a x + b}^2} \equiv -\dfrac {2 a} {b^3 x} + \dfrac 1 {b^2 x^2} + \dfrac {2 a^2} {b^3 \paren {a x + b} } + \dfrac {a^2} {b^2 \paren {a x + b}^2}$
Proof
| \(\ds \dfrac 1 {x^2 \paren {a x + b}^2}\) | \(\equiv\) | \(\ds \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b} + \dfrac D {\paren {a x + b}^2}\) | ||||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A x \paren {a x + b}^2 + B \paren {a x + b}^2 + C x^2 \paren {a x + b} + D x^2\) | multiplying through by $x^2 \paren {a x + b}^2$ | |||||||||
| \(\ds \) | \(\equiv\) | \(\ds A a^2 x^3 + 2 A a b x^2 + A b^2 x\) | multiplying everything out | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds B a^2 x^2 + 2 B a b x + B b^2\) | (tedious though this is, it helps to | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C a x^3 + C b x^2 + D x^2\) | identify the equal indices) |
Setting $a x + b = 0$ in $(1)$:
| \(\ds a x + b\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -\frac b a\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D \paren {-\frac b a}^2\) | \(=\) | \(\ds 1\) | substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds \frac {a^2} {b^2}\) |
Equating constants in $(1)$:
| \(\ds 1\) | \(=\) | \(\ds B b^2\) | ||||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac 1 {b^2}\) |
Equating $1$st powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A b^2 + 2 B a b\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A b^2\) | \(=\) | \(\ds -\frac {2 a b} {b^2}\) | subtituting for $B$ from $(2)$ | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds -\frac {2 a} {b^3}\) |
Equating $3$rd powers of $x$:
| \(\ds 0\) | \(=\) | \(\ds A a^2 + C a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac {2 a^2} {b^3}\) | substituting for $A$ from $(3)$ and simplifying |
Summarising:
| \(\ds A\) | \(=\) | \(\ds -\frac {2 a} {b^3}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \frac 1 {b^2}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac {2 a^2} {b^3}\) | ||||||||||||
| \(\ds D\) | \(=\) | \(\ds \frac {a^2} {b^2}\) |
Hence the result.
$\blacksquare$