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

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

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

Proof

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

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$

Substituting for $a$, $b$ and $c$ and simplifying:

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

\(=\)

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

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

\(\ds \map \ln {x + 2 + \sqrt {\paren {x + 2}^2 + 1} } + C\)

Definition 2 of Inverse Hyperbolic Sine

\(\ds \)

\(=\)

\(\ds \map \ln {x + 2 + \sqrt {x^2 + 4 x + 5} } + C\)

simplification

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $16$.