定积分上限为2下限为0 x/(x^2-2x+2)^2dx
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定积分上限为2下限为0 x/(x^2-2x+2)^2dx
答案是pi/4+1/2.我的感觉是令x-1=tanx,但就是不对算出来,
答案是pi/4+1/2.我的感觉是令x-1=tanx,但就是不对算出来,
N = ∫(0~2) x/[x² - 2x + 2]² dx
= ∫(0~2) x/[(x - 1)² + 1]² dx
Let x - 1 = tanz,dx = sec²z dz
N = ∫(- π/4~π/4) [1 + tanz]/[1 + tan²z]² * [sec²z dz]
= ∫(- π/4~π/4) [1 + tanz]/sec⁴z * [sec²z dz]
= ∫(- π/4~π/4) [1 + tanz][cos²z] dz
= ∫(- π/4~π/4) [cos²z + sinzcosz] dz
= ∫(- π/4~π/4) [1/2 + (1/2)cos2z + (1/2)sin2z] dz
= [z/2 + (1/2)sinzcosz - (1/4)cos2z] |(- π/4~π/4)
= [1/4 + π/8] - [- 1/4 - π/8]
= 1/2 + π/4
= ∫(0~2) x/[(x - 1)² + 1]² dx
Let x - 1 = tanz,dx = sec²z dz
N = ∫(- π/4~π/4) [1 + tanz]/[1 + tan²z]² * [sec²z dz]
= ∫(- π/4~π/4) [1 + tanz]/sec⁴z * [sec²z dz]
= ∫(- π/4~π/4) [1 + tanz][cos²z] dz
= ∫(- π/4~π/4) [cos²z + sinzcosz] dz
= ∫(- π/4~π/4) [1/2 + (1/2)cos2z + (1/2)sin2z] dz
= [z/2 + (1/2)sinzcosz - (1/4)cos2z] |(- π/4~π/4)
= [1/4 + π/8] - [- 1/4 - π/8]
= 1/2 + π/4
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