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Question

如题所述

Answer 1

电影

www.4K7BYVrE.cn

= π/4 + (1/2)∫[1，1] 1/u du

=lim(3x^2)/x^2=3

=sinx|（0->π/4)+cosx|（0->π/4)-cosx|（π/4->π/2)-sinx|（π/4->π/2)

令t=1/u 变换(2)式得 f(1/x)= ∫ln(1/u)/(1+(1/u )^2) d(1/u ) u= 1→x

∫[1-2](t(1-t)×2t)dt

=lim(x->1)(lnx+1)/[lnx+1+1]

=[2+1/(x-1)+1/(x-1)]/{x[1+/(x×x-1)]}

=e2-3

= (1/2)∫ dx + (1/2)∫ (cosx-sinx)/(sinx+cosx) dx，令u=sinx+cosx，du=(cosx-sinx)dx

Answer 2

手机

www.wOtdKPAn.cn

B=∫sinx/(sinx+cosx)dx

A(x^4+2x^2+1)+(Bx+C)(x^3-x^2+x-1)+(Dx^2+Ex-Dx-E)=1

=sinx+(1/2)∫sin2xdx

f(x)=∫[1,x] ln(1+t)/t dt 令u=1/t

=lim(x)[3cos^2(x)sinx]/2[1/(1+x)^2+1/(1+x)-2]

=∫（0-π/2）(1-sin2x)2·(sinx)2d(sinx)

=lim(3x^2)/x^2=3

a+b=1,a-b=0,a=b=1/2,

= (1/2)(lnt)^2 |[1→1/x]