Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2/Proof 1
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Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$
- $\ds \int \frac {\d x} {3 x^2 + 4 x + 2} = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C$
Proof
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
\(\ds a\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds 4\) | ||||||||||||
\(\ds c\) | \(=\) | \(\ds 2\) |
We note that:
\(\ds b^2 - 4 a c\) | \(=\) | \(\ds 4^2 - 4 \times 3 \times 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 16 - 24\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -8\) |
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
\(\ds \int \frac {\d x} {3 x^2 + 4 x + 2}\) | \(=\) | \(\ds \dfrac 2 {\sqrt {4 \times 3 \times 2 - 4^2} } \map \arctan {\dfrac {2 \times 3 x + 4} {\sqrt {4 \times 3 \times 2 - 4^2} } } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 {\sqrt 8} \map \arctan {\dfrac {6 x + 4} {\sqrt 8} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 {2 \sqrt 2} \map \arctan {\dfrac {2 \paren {3 x + 2} } {2 \sqrt 2} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C\) |
$\blacksquare$