参考にしたサイト➔置換積分法の公式やパターンを見分けるコツをわかりやすく解説 | 受験辞典

問題：次の式を計算せよ

• (1)の答え：\( \log(x + \sqrt{x^2 + 9}) + C \)

  解説：

  \( t = x + \sqrt{x^2 + 9} \)とおくと、
  \( \dfrac{dt}{dx} = 1 + \dfrac{2x}{2\sqrt{x^2 + 9}} \)より\( \dfrac{1}{\sqrt{x^2 + 9}}dx = \dfrac{1}{t}dt \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{1}{\sqrt{x^2 + 9}} dx & = & \displaystyle \int \dfrac{1}{t} dt \\ & = & \log|t| + C \\ & = & \log(x + \sqrt{x^2 + 9}) + C \\ \end{eqnarray} \)


• (2)の答え：\( 5\log(x + \sqrt{x^2 -4}) + C \)

  解説：

  \( t = x + \sqrt{x^2 - 4} \)とおくと、
  \( \dfrac{dt}{dx} = 1 + \dfrac{2x}{2\sqrt{x^2 - 4}} \)より\( \dfrac{1}{\sqrt{x^2 - 4}}dx = \dfrac{1}{t}dt \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{5}{\sqrt{x + 2}\sqrt{x - 2}} dx & = & 5\displaystyle \int \dfrac{1}{\sqrt{x^2 -4}} dx \\ & = & 5\displaystyle \int \dfrac{1}{t} dt \\ & = & 5\log|t| + C \\ & = & 5\log(x + \sqrt{x^2 -4}) + C \\ \end{eqnarray} \)


• (3)の答え：\( 7\log(3 + 2\sqrt{2}) \)

  解説：

  \( t = x + \sqrt{x^2 - 1} \)とおくと、
  \( \dfrac{dt}{dx} = 1 + \dfrac{2x}{2\sqrt{x^2 - 1}} \)より\( \dfrac{1}{\sqrt{x^2 - 1}}dx = \dfrac{1}{t}dt \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{1}^{3} \dfrac{7}{\sqrt{x^2 - 1}} dx & = & 7\displaystyle \int_{1}^{3 + 2\sqrt{2}} \dfrac{1}{t} dt \\ & = & 7\left[ \log|t| \right]_{1}^{3 + 2\sqrt{2}} \\ & = & 7(\log(3 + 2\sqrt{2}) - \log 1) \\ & = & 7\log(3 + 2\sqrt{2}) \end{eqnarray} \)


• (4)の答え：\( \dfrac{1}{2}\log(\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}) \)

  解説：

  \( x = \sin \theta \)とおくと、
  \( \dfrac{dx}{d\theta} = \cos \theta \)より\( dx = \cos \theta d\theta \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\frac{\sqrt{3}}{2}} \dfrac{1}{\sqrt{1 - x^2}} dx & = & \displaystyle \int_{0}^{\frac{\pi}{3}} \dfrac{1}{\sqrt{1 - \sin^2 \theta}} \cos \theta d\theta \\ & = & \displaystyle \int_{0}^{\frac{\pi}{3}} \dfrac{1}{\cos \theta} d\theta \\ & = & \left[ \dfrac{1}{2}\log(\dfrac{1 + \sin \theta}{1 - \sin \theta}) \right]_{0}^{\frac{\pi}{3}} \\ & = & \dfrac{1}{2}\log(\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}) \\ \end{eqnarray} \)


• (5)の答え：\( \dfrac{\pi}{4} \)

  解説：

  \( x = 5\tan \theta \)とおくと、
  \( \dfrac{dx}{d\theta} = \dfrac{5}{\cos^2 \theta} \)より\( dx = \dfrac{5}{\cos^2 \theta} d\theta \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{5} \dfrac{5}{x^2 + 25} dx & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{5}{25\tan^2 \theta + 25} \dfrac{5}{\cos^2 \theta} d\theta \\ & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{1}{\tan^2 \theta + 1} \dfrac{1}{\cos^2 \theta} d\theta \\ & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{1}{\dfrac{1}{\cos^2 \theta}} \dfrac{1}{\cos^2 \theta} d\theta \\ & = & \displaystyle \int_{0}^{\frac{\pi}{4}} d\theta \\ & = & \left[ \theta \right]_{0}^{\frac{\pi}{4}} \\ & = & \dfrac{\pi}{4} \end{eqnarray} \)