第30题,积分换元法,
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第30题,积分换元法,
第30题,积分换元法,
第30题,积分换元法,
there're 3 methods you could use.
actually the infininte has many solutions(non-unique),so you don't mind the forms of these solutions.each of them has the same derivative but differs from a constant C.
we could use the relationship arcsinx + arccosx = π/2 for interchange.
but one thing you still need to pay attention,these solutions are valid for the situation x > 1 only,for the situation x < 1,you should do once more substituion due to different domains of the inverse secant function.
设t-x^2=√(x^4-1),x=√((1+t^2)/2)
原积分=∫1/(√((1+t^2)/2)×(t-((1+t^2)/2)))dxdx=( √((1+t^2)/2))'=t/ √((1+t^2)/2原积分=∫t/((1+t^2)×(t-((1+t^2/2)修正一下，dx=t/√(2(1+t^2))原积分=∫-2t/((1+t^2)(1-t)^2)下面你自己解下这个多项式分式积分。...
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设t-x^2=√(x^4-1),x=√((1+t^2)/2)
原积分=∫1/(√((1+t^2)/2)×(t-((1+t^2)/2)))dx
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