積分(3)の1．(3)の解答

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 2x+1 }+\sqrt{ 2x } \hspace{2}}dx }$
$3$\displaystyle{\begin{array}{lll}=& \int \frac{\hspace{2} \sqrt{ 2x+1 }- \sqrt{ 2x } \hspace{2}}{\hspace{2} $ 2x+1 $ - 2x \hspace{2}}dx & \vspace{6}\\ =& \int \sqrt{ 2x+1 }- \sqrt{ 2x }dx & \vspace{6}\\ =& \int \{ { $ 2x+1 $ }^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }- { $ 2x $ }^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} } \} dx & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{ $ 2x+1 $ }^{ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{ $ 2x $ }^{ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} }+C & \vspace{6}\\ =& \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} $ 2x+1 $ \sqrt{ 2x+1 }- \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}x\sqrt{ 2x }+C & \vspace{6}\\ \end{array}}$

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