Question

calculati : $\int\limits^1_0 { \sqrt{x^2+1}-x } \, dx$

Answer 1

$\int _0^1\sqrt{x^2+1}-xdx \\ =\int \sqrt{x^2+1}dx-\int \:xdx \\ =\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln \left|x+\sqrt{1+x^2}\right|-\frac{x^2}{2} \\ =\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln \left|x+\sqrt{1+x^2}\right|-\frac{x^2}{2}+C \\ =-\frac{1}{2}+\frac{\ln \left(1+\sqrt{2}\right)+\sqrt{2}}{2}-0 \\ =\frac{\ln \left(1+\sqrt{2}\right)+\sqrt{2}-1}{2}$