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

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

$\ds \int \dfrac {\d x} {2 x + x^2} = $

Proof

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

\(\ds a\)

\(=\)

\(\ds 1\)

\(\ds b\)

\(=\)

\(\ds 2\)

\(\ds c\)

\(=\)

\(\ds 0\)

We note that:

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

\(=\)

\(\ds 2^2 - 4 \times 1 \times 0\)

\(\ds \)

\(=\)

\(\ds 4\)

Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:

$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$

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

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

\(=\)

\(\ds \dfrac 1 {\sqrt 1} \ln \size {2 \sqrt 1 \sqrt {1 \times x^2 + 2 \times x + 0} + 2 \times 1 \times x + 2} + C\)

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

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

Sum of Logarithms

\(\ds \)

\(=\)

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

subsuming $\ln 2$ into the arbitrary constant

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $20$.