∫dxx2+A\int \frac{dx}{\sqrt{x^2+A}}∫x2+Adx x2+A=t−x\sqrt{x^2+A} = t-xx2+A=t−x とおくと x2+A=(t−x)2=t2−2tx+x2x^2+A=(t-x)^2=t^2-2tx+x^2x2+A=(t−x)2=t2−2tx+x2 t2−2tx−A=0t^2-2tx-A=0t2−2tx−A=0 x=t2−A2tx=\dfrac{t^2-A}{2t}x=2tt2−A dx=2t⋅2t−(t2−A)⋅24t2dt=2t2+2A4t2dtdx=\dfrac{2t\cdot 2t-(t^2-A)\cdot 2}{4t^2}dt=\dfrac{2t^2+2A}{4t^2}dtdx=4t22t⋅2t−(t2−A)⋅2dt=4t22t2+2Adt これから =∫1t−t2−A2t⋅2t2+2A4t2dt=∫2t2t2−t2+A⋅t2+A2t2dt=∫1t2+A⋅t2+Atdt=∫1tdt=log∣t∣+C=\int \frac{1}{t-\dfrac{t^2-A}{2t}}\cdot\frac{2t^2+2A}{4t^2}dt=\int \frac{2t}{2t^2-t^2+A}\cdot\frac{t^2+A}{2t^2}dt=\int \frac{1}{t^2+A}\cdot\frac{t^2+A}{t}dt=\int \frac{1}{t}dt=\log|t|+C=∫t−2tt2−A1⋅4t22t2+2Adt=∫2t2−t2+A2t⋅2t2t2+Adt=∫t2+A1⋅tt2+Adt=∫t1dt=log∣t∣+C 元に戻して =log∣x+x2+A∣+C=\log\left|x+\sqrt{x^2+A}\right|+C=logx+x2+A+C (C は積分定数)