$3$\displaystyle{\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}}dx }}$
　 $3$\displaystyle{=\log \| x+\sqrt{ {x}^{2}+A } \| +C}$
　 $3$A$ は任意定数
　 $3$C$ は積分定数

$3$\displaystyle{\begin{array}{lll}{ \{ \log \| x+\sqrt{ {x}^{2}+A } \| \} }^{\prime }& \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}}{ $ x+\sqrt{ {x}^{2}+A } $ }^{\prime }& \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}} $ 1+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2} 2x \hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}} $ & \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}}\cdot \frac{\hspace{2} \sqrt{ {x}^{2}+A }+x \hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}}& \vspace{6}\\ \text{\hspace{5}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}}& \vspace{6}\\ \end{array}}$

■応用公式
$3$\displaystyle{\begin{array}{lll}\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ { $ ax+b $ }^{2}+A } \hspace{2}}dx }& \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}\log \| ax+b+\sqrt{ { $ ax+b $ }^{2}+A } \| & \vspace{6}\\ \text{\hspace{5}}+C& \vspace{6}\\ \end{array}}$
　 $3$a$ ， $3$b$ は定数

■計算例
$3$\displaystyle{\begin{array}{lll}\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4{x}^{2}+3 } \hspace{2}}dx }& \vspace{6}\\ =\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ { $ 2x $ }^{2}+3 } \hspace{2}}dx }& \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \| 2x+\sqrt{ { $ 2x $ }^{2}+3 } \| +C& \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \| 2x+\sqrt{ 4{x}^{2}+3 } \| +C& \vspace{6}\\ \end{array}}$