Primitive of Reciprocal of a x squared plus b x plus c/Examples/3 x^2 + 4 x + 2

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 1

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:

\(\ds \int \frac {\d x} {3 x^2 + 4 x + 2}\)

\(=\)

\(\ds \dfrac 2 {\sqrt {4 \times 3 \times 2 - 4^2} } \map \arctan {\dfrac {2 \times 3 x + 4} {\sqrt {4 \times 3 \times 2 - 4^2} } } + C\)

\(\ds \)

\(=\)

\(\ds \dfrac 2 {\sqrt 8} \map \arctan {\dfrac {6 x + 4} {\sqrt 8} } + C\)

\(\ds \)

\(=\)

\(\ds \dfrac 2 {2 \sqrt 2} \map \arctan {\dfrac {2 \paren {3 x + 2} } {2 \sqrt 2} } + C\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C\)

$\blacksquare$

Proof 2

\(\ds \int \frac {\d x} {3 x^2 + 4 x + 2}\)

\(=\)

\(\ds \dfrac 1 3 \int \frac {\d x} {x^2 + \frac 4 3 x + \frac 2 3}\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac 2 3 - \frac 4 9} }\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \frac 2 9}\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac {\sqrt 2} 3}^2}\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 3 \paren {\dfrac 3 {\sqrt 2} \map \arctan {\dfrac {x + \frac 2 3} {\frac {\sqrt 2} 3} } } + C\)

Primitive of $\dfrac 1 {x^2 + a^2}$

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C\)

$\blacksquare$