高等数学不定积分,求大神解答26.27.28题,高分悬赏
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发布时间：2024-01-23 20:26
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时间：2024-07-29 19:27
= ∫(x^3+x-x+1)dx/(1+x^2)^2
= ∫xdx/(1+x^2) - ∫xdx/(1+x^2)^2 + ∫dx/(1+x^2)^2
= (1/2)∫d(x^2+1)/(1+x^2) - (1/2)∫d(x^2+1)/(1+x^2)^2
+ ∫dx/(1+x^2)^2
= (1/2)ln(1+x^2) + 1/(1+x^2) + ∫dx/(1+x^2)^2
后者 令 x = tant,
I3 = ∫dx/(1+x^2)^2 = ∫dt/(sect)^2 = ∫(cost)^2dt
=(1/2) ∫(1+cos2t)dt = t/2+(1/4)sin2t+C
= (1/2)arctanx + 2x/(1+x^2) + C
则 I = (1/2)ln(1+x^2) + 1/(1+x^2) + (1/2)arctanx + 2x/(1+x^2) + C
27. 令 x = 3sect,
I = ∫√(x^2-9)dx/x = ∫3(tant)^2dt = 3∫[(sect)^2-1]dt
= 3tant - 3t + C = √(x^2-9) - 3arccos(x/3) + C
28. 令 x = 1/t,
I = ∫dx/[x(x^n+1)] = -∫t^(n-1)dt/(1+t^n)
= (-1/n)∫d(1+t^n)/(1+t^n) = (-1/n)ln|1+t^n|+C
= (-1/n)ln|(1+x^n)/x^n|+C
= ln|x| - (1/n)ln|(1+x^n)|+C
