Primitive of Reciprocal of x by Root of a x squared plus b x plus c/Proof 2

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\displaystyle \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = \begin{cases} \dfrac {-1} {\sqrt c} \ln \left({\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x}\right) & : b^2 - 4 a c > 0 \\ \dfrac {-1} {\sqrt c} \sinh^{-1} \left({\dfrac {b x + 2 c} {\left\vert{x}\right\vert \sqrt {4 a c - b^2} } }\right) & : b^2 - 4 a c < 0 \\ \dfrac {-1} {\sqrt c} \ln \left\vert{\dfrac {2 c} x + b}\right\vert + C & : b^2 - 4 a c = 0 \end{cases}$


Proof

\(\displaystyle x \sqrt {a x^2 + b x + c}\) \(=\) \(\displaystyle \frac x {\paren {a x^2 + b x + c}^{-\frac 1 2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {\paren {a x^2 + b x + c}^{-\frac 1 2} \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac N D\)
where:
\(\displaystyle N\) \(=\) \(\displaystyle \frac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x\)
\(\displaystyle D\) \(=\) \(\displaystyle \frac {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}\)

Then:

\(\displaystyle \frac {\d N} {\d x}\) \(=\) \(\displaystyle \frac {x \paren {\sqrt c \paren {2 a x + b} \paren {a x^2 + b x + c}^{-\frac 1 2} + b} - \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x} + b x \sqrt {a x^2 + b x + c} } - \paren {2 \sqrt c \paren {a x^2 + b x + c} + \paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x - 2 a x^2 - 2 b x - 2 c} + b x \sqrt {a x^2 + b x + c} } - \paren {\paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {-b x - 2 c} - \paren {\paren {2 c} \sqrt {a x^2 + b x + c} } } } {x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {2 c + \sqrt c \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt c \times D\)


Now provided $N$ is real and non zero:

\(\displaystyle \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) \(=\) \(\displaystyle \int \frac D N \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\sqrt c} \int \frac {\frac {\d N} {\d x} } N \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\sqrt c} \ln \size N\) Primitive of Function under its Derivative
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\sqrt c} \ln \size {\frac {\paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } x}\)