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

## 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 $\displaystyle$ $=$ $\displaystyle \frac {-1} {\sqrt c} \ln \size {\frac {\paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } x}$