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

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

$\ds \int \frac {\d x} {x^2 + 2 a x + b} = \dfrac 1 {\sqrt {b - a^2} } \map \arctan {\dfrac {x + a} {\sqrt {b - a^2} } } + C$

where $b > a^2$.

Proof

We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:

\(\ds a\)

\(\gets\)

\(\ds 1\)

\(\ds b\)

\(\gets\)

\(\ds 2 a\)

\(\ds c\)

\(\gets\)

\(\ds b\)

We note that:

\(\ds b\)

\(>\)

\(\ds a^2\)

\(\ds \leadsto \ \ \)

\(\ds \paren {2 a}^2 - 4 \times 1 \times b\)

\(<\)

\(\ds 0\)

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 $c$ by hypothesis:

\(\ds \int \frac {\d x} {x^2 + 2 a x + b}\)

\(=\)

\(\ds \dfrac 2 {\sqrt {4 \times 1 \times b - \paren {2 a}^2} } \map \arctan {\dfrac {2 \times 1 \times x + 2 a} {\sqrt {4 \times 1 \times b - \paren {2 a}^2} } } + C\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\sqrt {b - a^2} } \map \arctan {\dfrac {x + a} {\sqrt {b - a^2} } } + C\)

$\blacksquare$

Sources

• 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals