高等数学吧 关注：471,944贴子：3,875,159
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∫(0->1) f(x)/√x dx
=2∫(0->1) f(x) d√x
=2[f(x).√x]|(0->1) - 2∫(0->1) f'(x)√x dx
=0 -2∫(0->1) √x . [ln(x+1)/x] dx
=-2∫(0->1) [ln(x+1)/√x] dx
=-4∫(0->1) [ln(x+1) d√x]
=-4[√x.ln(x+1)]|(0->1) +4∫(0->1) √x/(x+1) dx
=-4ln2 +4∫(0->1) √x/(x+1) dx
u=√x
=-4ln2 +4∫(0->1) [u/(u^2+1)] (2udu)
=-4ln2 +8∫(0->1) u^2/(u^2+1) du
=-4ln2 +8∫(0->1) [1-1/(u^2+1)] du
=-4ln2 +8[u-arctanu]|(0->1)
=-4ln2 +8(1-π/4)
=8-4ln2 -2π
∫(0->1) f(x)/√x dx
=2∫(0->1) f(x) d√x
=2[f(x).√x]|(0->1) - 2∫(0->1) f'(x)√x dx
=0 -2∫(0->1) √x . [ln(x+1)/x] dx
=-2∫(0->1) [ln(x+1)/√x] dx
=-4∫(0->1) [ln(x+1) d√x]
=-4[√x.ln(x+1)]|(0->1) +4∫(0->1) √x/(x+1) dx
=-4ln2 +4∫(0->1) √x/(x+1) dx
u=√x
=-4ln2 +4∫(0->1) [u/(u^2+1)] (2udu)
=-4ln2 +8∫(0->1) u^2/(u^2+1) du
=-4ln2 +8∫(0->1) [1-1/(u^2+1)] du
=-4ln2 +8[u-arctanu]|(0->1)
=-4ln2 +8(1-π/4)
=8-4ln2 -2π
