Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0/Proof 1
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Theorem
Let $a \in \R_{\ne 0}$.
Let $b = 0$.
Then:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C & : a c > 0 \\ \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C & : a c < 0 \\ \dfrac {-1} {a x} + C & : c = 0 \end {cases}$
Proof
First:
| \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {\d x} {a x^2 + c}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\d x} {x^2 + \frac c a}\) | Primitive of Constant Multiple of Function |
Let $a c > 0$.
Then $\dfrac c a > 0$ and:
| \(\ds \frac c a\) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac c a\) | \(=\) | \(\ds q^2\) | for some $q \in \R$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds \sqrt {\frac c a}\) |
Thus:
| \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \frac 1 a \int \frac {\d x} {x^2 + q^2}\) | Integration by Substitution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a q} \arctan \frac x q + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a \sqrt {\frac c a} } \arctan \frac x {\sqrt {\frac c a} } + C\) | substituting for $q$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C\) | simplifying |
$\Box$
Let $a c < 0$.
Then $\dfrac c a > 0$ and:
| \(\ds \frac c a\) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -\frac c a\) | \(=\) | \(\ds q^2\) | for some $q \in \R$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds \sqrt {-\frac c a}\) |
Thus:
| \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \frac 1 a \int \frac {\d x} {x^2 - q^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C\) | substituting for $q$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C\) | simplifying |
$\Box$
Let $c = 0$.
Then:
| \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {\d x} {a x^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {-1} {a x} + C\) | Primitive of Power |
$\blacksquare$