$$
\begin{eqnarray}
\int \sqrt{x^2+a^2} \mathrm{d}x
&=&\int \sqrt{a^2\tan^2{\left(\theta\right)}+a^2}\;a\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\;\cdots\;
x=a\tan{\left(\theta\right)},
\frac{\mathrm{d}x}{\mathrm{d}\theta}
=a\frac{\mathrm{d}}{\mathrm{d}\theta}\tan{\left(\theta\right)}
=a\frac{1}{\cos^2{\left(\theta\right)}},
\mathrm{d}x=a\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\\&=&a \int \sqrt{a^2\left\{\tan^2{\left(\theta\right)}+1\right\}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\\&=&a \int a\sqrt{\tan^2{\left(\theta\right)}+1}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\\&=&a^2 \int \sqrt{\frac{1}{\cos^2{\left(\theta\right)}}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\;\cdots\;\tan^2{\left(\theta\right)}+1
=\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1
=\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
=\frac{1}{\cos^2{\left(\theta\right)}}
\\&=&a^2 \int \frac{1}{\cos{\left(\theta\right)}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta
\\&=&a^2 \int \frac{1}{\cos^3{\left(\theta\right)}}\mathrm{d}\theta
\\&=&a^2 \left[\frac{1}{2}\left\{
\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}
+\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|}
\right\}+C_0
\right]
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/cossec_25.html}{\int \frac{1}{\cos^3{\left(\theta\right)}}\mathrm{d}\theta
=\frac{1}{2}\left\{
\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}
+\ln{\left|
\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}
\right|}
\right\}+C_0\;(C_0:積分定数)}
\\&=&\frac{a^2}{2}\left\{
\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}
+\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|}
+C_0
\right\}
\\&=&\frac{1}{2}\left\{
a^2\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}
+a^2\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|}
+a^2C_0
\right\}
\\&=&\frac{1}{2}\left\{
a^2\;\frac{1}{a^2}x\sqrt{x^2+a^2}
+a^2\ln{\left|
\frac{1}{a}\left(x+\sqrt{x^2+a^2}\right)
\right|}
+a^2C_0
\right\}
\;\cdots\;
\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}=\frac{1}{a^2}x\sqrt{x^2+a^2},
\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}=\frac{1}{a}\left(x+\sqrt{x^2+a^2}\right)
\\&=&\frac{1}{2}\left[
x\sqrt{x^2+a^2}
+a^2\left\{
\ln{\left|x+\sqrt{x^2+a^2}\right|}
-\ln{\left|a\right|}
\right\}
+a^2C_0
\right]
\;\cdots\;\ln{\frac{A}{B}}=\ln{A}-\ln{B}
\\&=&\frac{1}{2}\left\{
x\sqrt{x^2+a^2}
+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}
-a^2\ln{\left|a\right|}
+a^2C_0
\right\}
\\&=&\frac{1}{2}\left\{
x\sqrt{x^2+a^2}
+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}
\right\}
-\frac{a^2}{2}\ln{\left|a\right|}
+\frac{a^2}{2}C_0
\\&=&\frac{1}{2}\left\{
x\sqrt{x^2+a^2}
+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}
\right\}
+C\;\cdots\;C=-\frac{a^2}{2}\ln{\left|a\right|}+\frac{a^2}{2}C_0\;(C:積分定数)
\end{eqnarray}
$$
$$
\begin{eqnarray}
\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}
&=&\tan{\left(\theta\right)}\sqrt{\frac{1}{\cos^2{\left(\theta\right)}}}
\\&=&\tan{\left(\theta\right)}\sqrt{\tan^2{\left(\theta\right)}+1}
\;\cdots\;
\frac{1}{\cos^2{\left(\theta\right)}}
=\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
=\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1
=\tan^2{\left(\theta\right)}+1
\\&=&\frac{x}{a}\sqrt{\left(\frac{x}{a}\right)^2+1}
\;\cdots\;x=a\tan{\left(\theta\right)}より\tan{\left(\theta\right)}=\frac{x}{a}
\\&=&\frac{x}{a}\sqrt{\frac{x^2+a^2}{a^2}}
\\&=&\frac{x}{a}\sqrt{\frac{1}{a^2}\left(x^2+a^2\right)}
\\&=&\frac{x}{a}\frac{1}{a}\sqrt{x^2+a^2}
\\&=&\frac{1}{a^2}x\sqrt{x^2+a^2}
\end{eqnarray}
$$
$$
\begin{eqnarray}
\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}
&=&\tan{\left(\theta\right)}+\sqrt{\frac{1}{\cos^2{\left(\theta\right)}}}
\\&=&\tan{\left(\theta\right)}+\sqrt{\tan^2{\left(\theta\right)}+1}
\;\cdots\;
\frac{1}{\cos^2{\left(\theta\right)}}
=\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}
=\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1
=\tan^2{\left(\theta\right)}+1
\\&=&\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1}
\;\cdots\;x=a\tan{\left(\theta\right)}より\tan{\left(\theta\right)}=\frac{x}{a}
\\&=&\frac{x}{a}+\sqrt{\frac{x^2+a^2}{a^2}}
\\&=&\frac{x}{a}+\sqrt{\frac{1}{a^2}\left(x^2+a^2\right)}
\\&=&\frac{x}{a}+\frac{1}{a}\sqrt{x^2+a^2}
\\&=&\frac{1}{a}\left(x+\sqrt{x^2+a^2}\right)
\end{eqnarray}
$$
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