Primitive of Reciprocal of a x squared plus b x plus c/Examples/x^2 + 4 x + 5

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

$\ds \int \dfrac {\d x} {x^2 + 4 x + 5} = \map \arctan {x + 2} + C$

Proof

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

\(\ds a\)

\(=\)

\(\ds 1\)

\(\ds b\)

\(=\)

\(\ds 4\)

\(\ds c\)

\(=\)

\(\ds 5\)

We note that:

\(\ds b^2 - 4 a c\)

\(=\)

\(\ds 4^2 - 4 \times 1 \times 5\)

\(\ds \)

\(=\)

\(\ds 16 - 20\)

\(\ds \)

\(=\)

\(\ds -4\)

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:

\(\ds \int \frac {\d x} {x^2 + 4 x + 5}\)

\(=\)

\(\ds \dfrac 2 {\sqrt {4 \times 1 \times 5 - 4^2} } \map \arctan {\dfrac {2 \times 1 \cdot x + 4} {\sqrt {4 \times 1 \times 5 - 4^2} } } + C\)

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

\(\ds \map \arctan {x + 2} + C\)

simplifying

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $15$.