∫ln(x+√1+x^2)dx怎么求?∫范围是0-1√下是（1+x^2）ln2-√2+1请问谁会求?.

Question

∫ln(x+√1+x^2)dx 怎么求?∫范围是0-1
√下是（1+x^2）
ln2- √2+1
请问谁会求?.

Answer 1

∫(0,1)ln(x+√1+x^2)dx
=x*ln(x+√1+x^2)(0,1)-∫(0,1)xdln(x+√1+x^2)
=ln(1+√2)-∫(0,1)x/√(1+x^2)*dx
=ln(1+√2)-∫(0,π/4)tant/√(1+tant^2)*dtant
=ln(1+√2)-∫(0,π/4)tant*sectdt
=ln(1+√2)-sect|(0,π/4)
=ln(1+√2)-(√2-1)