$$ \displaystyle \int_{0}^{2} \frac{2x+1}{\sqrt{x^2+4}} dx $$
【解答】
\begin{array}{l}
I & = \displaystyle \int_{0}^{2} \frac{2x}{\sqrt{x^2+4}} dx + \int_{0}^{2} \frac{1}{\sqrt{x^2+4}} dx \\
I_1 & = \displaystyle \int_{0}^{2} \frac{2x}{\sqrt{x^2+4}} dx \\
I_2 & = \displaystyle \int_{0}^{2} \frac{1}{\sqrt{x^2+4}} dx \\
\end{array}
とおく。ここで、\( \sqrt{x^2+4}=t \cdots ① \) とおくと、
\begin{array}{c|ccc}
x & 0 & \to & 2 \\
\hline
t & 2 & \to & 2\sqrt{2}
\end{array}
また①より \( x^2+4=t^2 \)
①の両辺を \( x \) で微分すると \( 2xdx = 2tdt \)
だから
\begin{eqnarray}
I_1 & = & \displaystyle \int_{2}^{2\sqrt{2}} \frac{2tdt}{t} \\
& = & \displaystyle \int_{2}^{2\sqrt{2}}2dt \\
& = & \left[ 2t \right]_{2}^{2\sqrt{2}} \\
& = & 4 ( \sqrt{2} -1 )
\end{eqnarray}
次に、\(x= 2\tan \theta \cdots ② \) とおくと、
\begin{array}{c|ccc}
x & 0 & \to & 2 \\
\hline
\theta & 0 & \to & \frac{\pi}{4}
\end{array}
また②より
\begin{eqnarray}
\sqrt{x^2+4} & = & 2\sqrt{1+\tan^2 \theta} = 2\sqrt{\frac{1}{\cos^2 \theta}} \\
& = & \frac{2}{\cos \theta} \ \ (∵\cos \theta \gt 0)
\end{eqnarray}
②の両辺を \( x \) で微分すると、\( dx = \frac{2}{\cos^2 \theta}d \theta \)
だから
\begin{eqnarray}
I_2 & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{2} \cdot \frac{2}{\cos^2 \theta} d \theta \\
& = & \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} d \theta \\
\end{eqnarray}
ここで、
\begin{eqnarray}
\frac{1}{\cos \theta} & = & \frac{\cos \theta}{\cos^2 \theta} = \frac{\cos \theta}{1 – \sin^2 \theta} \\
& = & \frac{\cos \theta}{(1 – \sin \theta)(1 + \sin \theta) } \\
& = & \frac{1}{2} \left( \frac{\cos \theta}{1 – \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} \right) \\
\end{eqnarray}
だから、
\begin{eqnarray}
I_2 & = & \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \left( \frac{\cos x}{1 – \sin x} + \frac{\cos x}{1 + \sin x} \right)dx \\
& = & \frac{1}{2} \left[ – \log | 1 – \sin x | + \log | 1 + \sin x | \right]_{0}^{\frac{\pi}{4}} \\
& = & \frac{1}{2} \left( – \log \left| 1- \frac{1}{\sqrt{2}} \right| + \log \left| 1+ \frac{1}{\sqrt{2}} \right| \right) \\
& = & \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) \\
& = & \frac{1}{2} \log ( \sqrt{2} + 1 )^{2} \\
& = & \log ( \sqrt{2} + 1 )
\end{eqnarray}
よって、
\begin{eqnarray}
I & = & I_1 + I_2 \\
& = & 4 ( \sqrt{2} -1 ) + \log ( \sqrt{2} + 1 ) \ \cdots （答）
\end{eqnarray}
\(I_2\)は以下のように置換すると、簡単に計算できます。
\( x + \sqrt{x^2+4} = t \ \cdots ③\) とおく。③の両辺を \(x\) で微分すると、
\begin{eqnarray}
\left( 1+\frac{2x}{2\sqrt{x^2+4}} \right) dx & = & dt \\
\frac{x+\sqrt{x^2+4}}{\sqrt{x^2+4}}dx & = & dt \\
\frac{1}{\sqrt{x^2+4}}dx & = & \frac{1}{x+\sqrt{x^2+4}} dt \\
\frac{1}{\sqrt{x^2+4}}dx & = & \frac{1}{t}dt
\end{eqnarray}
ここで、
\begin{array}{c|ccc}
x & 0 & \to & 2 \\
\hline
t & 2 & \to & 2 + 2\sqrt{2}
\end{array}
だから、
\begin{eqnarray}
I_2 & = & \displaystyle \int_{2}^{2+2\sqrt{2}} \frac{1}{t} dt \\
& = & \left[ \log | t | \right]_{2}^{2+2\sqrt{2}} \\
& = & \log \left( \frac{2+2\sqrt{2}}{2} \right) \\
& = & \log (\sqrt{2}+1)
\end{eqnarray}
