分母に1乗と2乗ルートの積分

分母に1乗と2乗ルートの積分

(1)

\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz=\frac{\sqrt{z^{2}-1}}{\pm z+1}+C
]

(2)

\begin{align*} \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz & =-\sqrt{2}\tanh^{\circ}\pm\frac{\pm z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C\\ & =-\frac{1}{\sqrt{2}}\tanh^{\circ}\left(\frac{1\mp z}{\sqrt{2}\sqrt{z^{2}+1}}\right) \end{align*}

(1)

\begin{align*} \int\frac{1}{\left(z+1\right)\sqrt{z^{2}-1}}dz & =\int\frac{1}{\left(\cosh t+1\right)\sqrt{\cosh^{2}t-1}}\sinh tdt\cnd{z=\cosh t}\\ & =\int\frac{1}{\left(\cosh t+1\right)\sqrt{\sinh^{2}t}}\sinh tdt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\int\frac{1}{\cosh t+1}dt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\int\frac{1}{2\cosh^{2}\frac{t}{2}}dt\\ & =\frac{\sinh t}{\sqrt{\sinh^{2}t}}\tanh\frac{t}{2}+C\\ & =\frac{2\sinh^{2}\frac{t}{2}}{\sqrt{\sinh^{2}t}}+C\\ & =\frac{\cosh t-1}{\sqrt{\sinh^{2}t}}+C\\ & =\frac{z-1}{\sqrt{\sinh^{2}\cosh^{\circ}z}}+C\\ & =\frac{z-1}{\sqrt{\left(\sqrt{\frac{z-1}{z+1}}\left(z+1\right)\right)^{2}}}+C\\ & =\frac{z-1}{\sqrt{\left(z-1\right)\left(z+1\right)}}+C\\ & =\frac{\left(z-1\right)\left(z+1\right)}{\left(z+1\right)\sqrt{\left(z-1\right)\left(z+1\right)}}+C\\ & =\frac{\sqrt{z^{2}-1}}{z+1}+C \end{align*}

\begin{align*} \int\frac{1}{\left(z-1\right)\sqrt{z^{2}-1}}dz & =\int\frac{1}{\left(\left(-z\right)+1\right)\sqrt{\left(-z\right)^{2}-1}}d\left(-z\right)\\ & =\frac{\sqrt{\left(-z\right)^{2}-1}}{\left(-z\right)+1}+C\\ & =\frac{\sqrt{z^{2}-1}}{-z+1}+C \end{align*}

これより、

\begin{align*} \int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz & =\frac{\sqrt{z^{2}-1}}{\pm z+1}+C \end{align*}

(2)

\begin{align*} \int\frac{1}{\left(z+1\right)\sqrt{z^{2}+1}}dz & =\int\frac{1}{\left(\tan t+1\right)\sqrt{\tan^{2}t+1}}\frac{1}{\cos^{2}t}dt\cnd{z=\tan t}\\ & =\frac{\sqrt{\cos^{2}t}}{\cos t}\int\frac{1}{\left(\tan t+1\right)\cos t}dt\\ & =\frac{\sqrt{\cos^{2}t}}{\cos t}\int\frac{1}{\left(\sin t+\cos t\right)}dt\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\circ}\frac{1-\tan\frac{t}{2}}{\sqrt{2}}+C\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\circ}\frac{1-\frac{\sin t}{1+\cos t}}{\sqrt{2}}+C\\ & =-\frac{\sqrt{\cos^{2}t}}{\cos t}\frac{2}{\sqrt{2}}\tanh^{\circ}\frac{1+\cos t-\sin t}{\sqrt{2}\left(1+\cos t\right)}+C\\ & =-\sqrt{2}\tanh^{\circ}\frac{\frac{1-z}{\sqrt{z^{2}+1}}+1}{\sqrt{2}\left(\frac{1}{\sqrt{z^{2}+1}}+1\right)}+C\\ & =-\sqrt{2}\tanh^{\circ}\frac{1+\sqrt{z^{2}+1}-z}{\sqrt{2}\left(1+\sqrt{z^{2}+1}\right)}+C\\ & =-\sqrt{2}\tanh^{\circ}\frac{-z^{2}-z\left(1-\sqrt{z^{2}+1}\right)}{\sqrt{2}\left(-z^{2}\right)}+C\\ & =-\sqrt{2}\tanh^{\circ}\frac{z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C \end{align*}

または、

\[
\int\frac{1}{\left(z+1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\circ}\left(\frac{1-z}{\sqrt{2}\sqrt{z^{2}+1}}\right)
]

でもよい(略)。

\begin{align*} \int\frac{1}{\left(z-1\right)\sqrt{z^{2}+1}}dz & =\int\frac{1}{\left(\left(-z\right)+1\right)\sqrt{\left(-z\right)^{2}+1}}d\left(-z\right)\\ & =-\sqrt{2}\tanh^{\circ}\frac{\left(-z\right)+1-\sqrt{\left(-z\right)^{2}+1}}{\sqrt{2}\left(-z\right)}+C\\ & =-\sqrt{2}\tanh^{\circ}-\frac{-z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C \end{align*}

または、

\[
\int\frac{1}{\left(z-1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\circ}\left(\frac{1+z}{\sqrt{2}\sqrt{z^{2}+1}}\right)
]

でもよい(略)。

これより、

\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz=-\sqrt{2}\tanh^{\circ}\pm\frac{\pm z+1-\sqrt{z^{2}+1}}{\sqrt{2}z}+C
]

または(略)、

\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}+1}}dz=-\frac{1}{\sqrt{2}}\tanh^{\circ}\left(\frac{1\mp z}{\sqrt{2}\sqrt{z^{2}+1}}\right)
]

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タイトル

分母に1乗と2乗ルートの積分

URL

https://www.nomuramath.com/s119x15e/

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