不定積分の問題

■別解

${\displaystyle \int \frac{1}{\sqrt{x^2+5}}dx}$ ${\displaystyle =\int \frac{1}{\sqrt{5}\sqrt{{\left(\frac{x}{\sqrt{5}}\right)}^2+1}}dx}$

${\displaystyle \frac{x}{\sqrt{5}}=\tan t}$ （ ${\displaystyle -\frac{1}{2}\pi <t<\frac{1}{2}\pi }$ ）とおき置換積分をする．

${\displaystyle x=\sqrt{5}\tan t}$ ， ${\displaystyle \frac{dx}{dt}=\frac{\sqrt{5}}{{\cos }^{2}t}}$ ， ${\displaystyle dx=\frac{\sqrt{5}}{{\cos }^{2}t}dt}$ ${\displaystyle \tan t}$ の微分はここを参照

よって

与式＝ ${\displaystyle {\displaystyle \int \frac{1}{\sqrt{5}\sqrt{{\tan }^{2}t+1}}\cdot \frac{\sqrt{5}}{{\cos }^{2}t}dt}}$

${\displaystyle ={\displaystyle \int \frac{1}{\sqrt{5}\sqrt{\frac{1}{{\cos }^{2}t}}}\cdot \frac{\sqrt{5}}{{\cos }^{2}t}dt}}$

${\displaystyle -\frac{1}{2}\pi <x<\frac{1}{2}\pi }$ では ${\displaystyle \cos x>0}$ より，簡単に√を取り除くことができる．

${\displaystyle ={\displaystyle \int \frac{1}{\sqrt{5}\frac{1}{\cos t}}\cdot \frac{\sqrt{5}}{{\cos }^{2}t}dt}}$

${\displaystyle ={\displaystyle \int \frac{1}{\cos t}dt}}$

${\displaystyle =\frac{1}{2}\log \left(\frac{1+\sin t}{1-\sin t}\right)+C}$ この計算はここを参照

${\displaystyle =\frac{1}{2}\log \frac{{\left(\sqrt{x^2+5}+x\right)}^2}{5}+C}$

${\displaystyle =\frac{1}{2}\log {\left(\sqrt{x^2+5}+x\right)}^2+\frac{1}{2}\log 5+C}$

${\displaystyle =\log \left(\sqrt{x^2+5}+x\right)+C^{\prime }}$ この対数の計算公式を参照 　･･････(3)

ただし， ${\displaystyle C^{\prime }=\frac{1}{2}\log 5+C}$ とおいている．

${\displaystyle \sqrt{x^2+5}+x>0}$ より，絶対値をとらなくてもよい．