设f(X)的导数=arctan(x-1)^2,f(0)=0,求不定积分(0,1)f(X)dx求详解
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设f(X)的导数=arctan(x-1)^2,f(0)=0,求不定积分(0,1)f(X)dx求详解
∫(0,1)f(X)dx
= ∫(0,1)f(X)d(x-1)
=(x-1)f(x)|(0,1) - ∫(0,1)(x-1)f‘(X)dx
= - ∫(0,1)(x-1)arctan(x-1)^2dx (u=(x-1)^2
=(1/2) ∫(0,1)arctanudu
=(1/2)(uarctanu|(0,1)-∫(0,1)u^2/(1+u^2)du)
=(1/2)(uarctanu-u+arctanu)|(0,1)
=(1/2)(π/2-1)
= ∫(0,1)f(X)d(x-1)
=(x-1)f(x)|(0,1) - ∫(0,1)(x-1)f‘(X)dx
= - ∫(0,1)(x-1)arctan(x-1)^2dx (u=(x-1)^2
=(1/2) ∫(0,1)arctanudu
=(1/2)(uarctanu|(0,1)-∫(0,1)u^2/(1+u^2)du)
=(1/2)(uarctanu-u+arctanu)|(0,1)
=(1/2)(π/2-1)
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