(*)分母に1乗と2乗ルートの積分
\[
\int\frac{1}{\left(z+\alpha\right)\sqrt{z^{2}+\beta}}dz=\frac{\tanh^{\circ}\left(\frac{\alpha z-\beta}{\sqrt{\alpha^{2}+\beta}\sqrt{\beta+z^{2}}}\right)}{\sqrt{\alpha^{2}+\beta}}
\]
分母に2乗のルートがある積分
\[
\int\frac{1}{\sqrt{z^{2}+\alpha}}dz=\frac{\sqrt{\alpha}\sqrt{\frac{z^{2}}{\alpha}+1}}{\sqrt{z^{2}+\alpha}}\sinh^{\circ}\frac{z}{\sqrt{\alpha}}+C
\]
分母に1乗と2乗ルートの積分
\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz=\frac{\sqrt{z^{2}-1}}{\pm z+1}+C
\]
三角関数・双曲線関数の一次結合の逆数の積分
\[
\int\frac{1}{\alpha\sin z+\beta\cos z+\gamma}dz=-\frac{2}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}\tanh^{\circ}\frac{\left(\gamma-\beta\right)\tan\frac{z}{2}+\alpha}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}+C
\]
三角関数(双曲線関数)の逆三角関数(逆双曲線関数)が恒等写像になる条件
\[
\sin^{\circ}\sin z=?z
\]
ヘヴィサイドの階段関数の問題
\[
f\left(H\left(\pm_{1}1\right)\right)g\left(-H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)g\left(H\left(\mp_{1}1\right)\right)=\left\{ f\left(0\right)g\left(0\right)+f\left(\pm1\right)g\left(\mp1\right)\right\} H\left(\pm_{2}1\right)\mp_{1}\left\{ f\left(0\right)g\left(0\right)-f\left(\pm_{1}1\right)g\left(\mp_{1}1\right)\right\} H\left(\mp_{2}1\right)
\]
ヘヴィサイドの階段関数の２定義値を引数に持つ関数の和と差
\[
f\left(H\left(\pm_{1}1\right)\right)\pm_{2}f\left(-H\left(\mp_{1}1\right)\right)=\left(f\left(0\right)+f\left(\pm_{1}1\right)\right)H\left(\pm_{2}1\right)\mp_{1}\left(f\left(0\right)-f\left(\pm_{1}1\right)\right)H\left(\mp_{2}1\right)
\]
ヘヴィサイドの階段関数の２定義値の和と差
\[
H\left(\pm_{1}1\right)\pm_{2}H\left(\pm_{1}1\right)=H\left(\pm_{2}1\right)\pm_{1}H\left(\pm_{2}1\right)
\]
ヘヴィサイドの階段関数の２定義値と関数
\[
f\left(x\right)H\left(\pm1\right)=f\left(\pm x\right)H\left(\pm1\right)
\]
ヘヴィサイドの階段関数の２定義値と複号
\[
H\left(\pm1\right)=\frac{1\pm1}{2}
\]