Question

定积分这道题？

第五题

Answer 1

f(x)=xcosx/(1+x^2)

f(-x) = -f(x)

=>∫(-π/2->π/2) xcosx/(1+x^2)  dx =0

∫(-π/2->π/2) [ (sinx)^2 +xcosx/(1+x^2) ] dx

=∫(-π/2->π/2)  (sinx)^2 dx

=2∫(0->π/2)  (sinx)^2 dx

=∫(0->π/2)  [1-cos2x ] dx

=[ x-(1/2)sin2x]|(0->π/2)

=π/2

Answer 2

额，你的图片实在是看不清，希望能解你的燃眉之急…………

Answer 3

设1/[x(x-2)²] =a/x+b/(x-2)+c/(x-2)²
则1=a(x-2)²+bx(x-2)+cx
=(a+b)x²+(-4a-2b+c)x+4a
故a+b=0，-4a-2b+c=0，4a=1，
得a=1/4，b=-1/4，c=1/2,
故原式=1/4 ∫dx/x - 1/4 ∫dx/(x-2) +½ ∫dx/(x-2)²
=1/4 ln|x| - 1/4 ln|x-2| -1/[2(x-2)]+c