By quotient rule, \begin {align*} f'(x) & = \frac {\sqrt {x^2 + p} - x \cdot \frac {1}{2} \cdot 2x \cdot \frac {1}{\sqrt {x^2 + p}}}{x^2 + p} \\ & = \frac {\sqrt {x^2 + p} - \frac {x^2}{\sqrt {x^2 + p}}}{x^2 + p} \\ & = \frac {p}{(x^2 + p) \sqrt {x^2 + p}}. \end {align*}
This gives \[ 0 < f'(x) \leq \frac {1}{\sqrt {p}}, \] with the equal sign taking if and only if \(x = 0\).
\(\lim _{x \to \infty } f(x) = 1\), so \(y = 1\) is a horizontal asymptote to the function.
Hence, the graph looks as follows:
Since \(y = \frac {cx}{\sqrt {x^2 + p}} = c f(x)\), we have \[ \DiffFrac {y}{x} = c f'(x) = \frac {cp}{\left (\sqrt {x^2 + p}\right )^3}, \] and hence \[ \Diff y = \frac {cp}{\left (\sqrt {x^2 + p}\right )^3} \Diff x. \]
The integral can therefore be simplified as \begin {align*} I & = \int \frac {\Diff y}{(b^2 - y^2) \sqrt {c^2 - y^2}} \\ & = \int \frac {1}{\left (b^2 - \frac {c^2 x^2}{x^2 + p}\right ) \sqrt {c^2 - \frac {c^2 x^2}{x^2 + p}}} \cdot \frac {cp}{\left (\sqrt {x^2 + p}\right )^3} \Diff x \\ & = \int \frac {cp \Diff x}{\left (b^2(x^2 + p) - c^2 x^2\right ) \sqrt {c^2 (x^2 + p) - c^2 x^2}} \\ & = \int \frac {cp \Diff x}{\left [(b^2 - c^2)x^2 + b^2 p\right ]\sqrt {c^2 p}} \\ & = \int \frac {\sqrt {p} \Diff x}{b^2 p + (b^2 - c^2) x^2}. \end {align*}
Let \(p = 1\), and we have \[ I = \int \frac {\Diff x}{b^2 + (b^2 - c^2)x^2} \] as desired.
Hence, \begin {align*} I & = \int \frac {\Diff x}{b^2 + (b^2 - c^2)x^2} \\ & = \frac {1}{b^2 - c^2} \int \frac {\Diff x}{\left (\frac {b}{\sqrt {b^2 - c^2}}\right )^2 + x^2} \\ & = \frac {1}{b^2 - c^2} \cdot \frac {\sqrt {b^2 - c^2}}{b} \arctan \frac {\sqrt {b^2 - c^2}x}{b} + C \\ & = \frac {1}{b \sqrt {b^2 - c^2}} \arctan \frac {\sqrt {b^2 - c^2}x}{b} + C. \end {align*}
Let \(b = \sqrt {3}\) and \(c = \sqrt {2}\), and hence \[ I = \frac {1}{\sqrt {3} \sqrt {3 - 2}} \arctan \frac {\sqrt {3 - 2}x}{\sqrt {3}} + C = \frac {1}{\sqrt {3}} \arctan \frac {x}{\sqrt {3}} + C. \]
When \(y = 1\), \(\frac {\sqrt {2} x}{\sqrt {x^2 + 1}} = 1\), and hence \(x^2 + 1 = 2x^2\), \(x^2 = 1\), giving \(x = 1\).
When \(y \to \sqrt {2} = b\), \(x \to \infty \).
Hence, \begin {align*} \int _{1}^{\sqrt {2}} \frac {\Diff y}{(3 - y^2) \sqrt {2 - y^2}} & = \frac {1}{\sqrt {3}} \left [\arctan \frac {x}{\sqrt {3}}\right ]_{1}^{\infty } = \frac {1}{\sqrt {3}} \left (\frac {\pi }{2} - \frac {\pi }{6}\right ) = \frac {\pi }{3\sqrt {3}}. \end {align*}
Consider letting \(x = \frac {1}{y}\) in the integral, and we have \(\Diff x = -\frac {1}{y^2} \Diff y = -x^2 \Diff y\), and when \(y = 1\), \(x = 1\), and when \(y = \sqrt {2}\), \(x = \frac {1}{\sqrt {2}}\). Hence, \begin {align*} \int _{\frac {1}{\sqrt {2}}}^{1} \frac {y \Diff y}{(3y^2 - 1)\sqrt {2 y^2 - 1}} & = \int _{\sqrt {2}}^{1} \frac {\frac {1}{x} \cdot \frac {1}{-x^2} \Diff x}{\left (\frac {3}{x^2} - 1\right ) \sqrt {\frac {2}{x^2} - 1}} \\ & = \int _{1}^{\sqrt {2}} \frac {\Diff x}{\left (3 - x^2\right )\sqrt {2 - x^2}} \\ & = \frac {\pi }{3\sqrt {3}}. \end {align*}
Consider the same substitution \(y = \frac {ax}{\sqrt {x^2 + p}}\). We still have \[ \Diff y = \frac {ap}{\left (\sqrt {x^2 + p}\right )^3} \Diff x, \] and hence \begin {align*} & \phantom {=} \int \frac {\Diff y}{(3y^2 - 1) \sqrt {2y^2 - 1}} \\ & = \int \frac {ap}{\left (\sqrt {x^2 + p}\right )^3} \cdot \frac {\Diff x}{\left (3 \cdot \frac {a^2 x^2}{x^2 + p} - 1\right ) \sqrt {2 \cdot \frac {a^2 x^2}{x^2 + p} - 1}} \\ & = \int \frac {ap \Diff x}{\left (3 a^2 x^2 - (x^2 + p)\right ) \sqrt {2 a^2 x^2 - (x^2 + p)}} \\ & = \int \frac {ap \Diff x}{\left ((3 a^2 - 1) x^2 - p\right ) \sqrt {(2 a^2 - 1) x^2 - p}}. \end {align*}
Consider letting \(a = \frac {1}{\sqrt {2}}\) and \(p = -1\), and we have \begin {align*} & \phantom {=} \int \frac {\Diff y}{(3y^2 - 1) \sqrt {2y^2 - 1}} \\ & = \int \frac {-\Diff x}{\sqrt {2} \left (\frac {1}{2} x^2 + 1\right )} \\ & = \int \frac {- \sqrt {2}\Diff x}{x^2 + 2} \\ & = -\sqrt {2} \cdot \frac {1}{\sqrt {2}} \arctan \frac {x}{\sqrt {2}} + C \\ & = - \arctan \frac {x}{\sqrt {2}} + C. \end {align*}
When \(y = \frac {1}{\sqrt {2}}\), we have \(\frac {1}{\sqrt {2}} \cdot \frac {x}{\sqrt {x^2 - 1}} = \frac {1}{\sqrt {2}}\), and \(x \to \infty \). When \(y = 1\), we have \(\frac {1}{\sqrt {2}} \cdot \frac {x}{\sqrt {x^2 - 1}} = 1\), and \(x = \sqrt {2}\). Hence, \begin {align*} & \phantom {=} \int _{\frac {1}{\sqrt {2}}}^1 \frac {\Diff y}{(3y^2 - 1) \sqrt {2y^2 - 1}} \\ & = -\left [\arctan \frac {x}{\sqrt {2}}\right ]_{\infty }^{\sqrt {2}} \\ & = \left [\arctan \frac {x}{\sqrt {2}}\right ]_{\sqrt {2}}^{\infty } \\ & = \frac {\pi }{2} - \frac {\pi }{4} \\ & = \frac {\pi }{4}. \end {align*}