∫√(x^2+A)dx

$ 　 \boxed{ ここで x = \frac{\sqrt{A}}{2}\left( \frac{t}{\sqrt{A}} - \frac{\sqrt{A}}{t} \right) より \frac{t}{\sqrt{A}} = e^u と置くと x = \sqrt{A}\sinh u となる。 }

$ \int \sqrt{x^2+A} dx = \int \underbrace{ \left(\frac t2 + \frac{A}{2t}\right) }_{(3)} \underbrace{ \left(\frac12 + \frac{A}{2t^2}\right)dt }_{(4)}

$ = \int \left(\frac{t}{4} + \frac{A}{2t} + \frac{A^2}{4t^3} \right)dt = \frac{t^2}{8} + \frac A2 \ln |t| - \frac{A^2}{8t^2} + C

$ = \frac12\left(\frac{t}{2}\right)^2 - \frac12\left(\frac{A}{2t}\right)^2 + \frac A2 \ln t + C = \frac12 \underbrace{ \left(\frac{t}{2}-\frac{A}{2t}\right) }_{(2)} \underbrace{ \left(\frac{t}{2}+\frac{A}{2t}\right) }_{(3)} + \frac12 \ln \underbrace{ |t| }_{(1)} + C

$ = \frac12 x \sqrt{ x^2 + 1 } + \frac12 \ln |x+\sqrt{x^2+A}| + C

$ \int_0^1 \sqrt{x^2+1} dx = \underbrace{ \int_1^{1+\sqrt2} }_{(1)'} \underbrace{ \left(\frac t2 + \frac{1}{2t}\right) }_{(3)} \underbrace{ \left(\frac12 + \frac{1}{2t^2}\right)dt }_{(4)}

$ = \int_1^{1+\sqrt2}\left(\frac{t}{4} + \frac{1}{2t} + \frac{1}{4t^3} \right)dt = \left[ \frac{t^2}{8} + \frac12 \ln t - \frac{1}{8t^2} \right]_1^{1+\sqrt2}

$ = \frac{(1+\sqrt2)^2 - 1^2}{8} + \frac{\ln (1+\sqrt2) - \ln 1}{2} - \frac18\left(\frac{1}{(1+\sqrt2)^2} - \frac1{1^2} \right)

$ = \frac18 \left( 3+2\sqrt2 - \frac{1}{3+2\sqrt2}\frac{3-2\sqrt2}{3-2\sqrt2} \right) + \frac12\ln (1+\sqrt2)

$ = \frac18 \left( 3+2\sqrt2 - 3+2\sqrt2 \right) + \frac12\ln (1+\sqrt2) = \frac{\sqrt2}{2} + \frac12\ln (1+\sqrt2)