∫ln(x+√1+x^2)dx 怎么求?∫范围是0-1
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∫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?.
√下是(1+x^2)
ln2- √2+1
请问谁会求?.
![∫ln(x+√1+x^2)dx 怎么求?∫范围是0-1](/uploads/image/z/17605961-17-1.jpg?t=%E2%88%ABln%28x%2B%E2%88%9A1%2Bx%5E2%29dx+%E6%80%8E%E4%B9%88%E6%B1%82%3F%E2%88%AB%E8%8C%83%E5%9B%B4%E6%98%AF0-1)
∫(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)
再问: 为什么 (ln(x+√1+x^2))'= 1/√(1+x^2)
再答: (ln(x+√1+x^2))' = 1/(x+√1+x^2)*(x+√1+x^2)' =1/(x+√1+x^2)*(1+x/√1+x^2) =1/√(1+x^2)
=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)
再问: 为什么 (ln(x+√1+x^2))'= 1/√(1+x^2)
再答: (ln(x+√1+x^2))' = 1/(x+√1+x^2)*(x+√1+x^2)' =1/(x+√1+x^2)*(1+x/√1+x^2) =1/√(1+x^2)