求下列定积分的过程
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（1）原式=-∫(0,π/4) xd(cosx)
=-xcosx|(0,π/4)+∫(0,π/4) cosxdx
=-π/4√2+sinx|(0,π/4)
=1/√2-π/4√2
（2）原式=xarctanx|(0,1)-∫(0,1) x/(1+x^2)dx
=π/4-(1/2)*ln|1+x^2||(0,1)
=π/4-(1/2)*ln2
（3）原式=∫(0,1) x^2d(e^x)
=x^2*e^x|(0,1)-∫(0,1) 2xe^xdx
=e-∫(0,1) 2xd(e^x)
=e-2xe^x|(0,1)+∫(0,1)2e^xdx
=e-2e+2e^x|(0,1)
=-e+2e-2
=e-2
=-xcosx|(0,π/4)+∫(0,π/4) cosxdx
=-π/4√2+sinx|(0,π/4)
=1/√2-π/4√2
（2）原式=xarctanx|(0,1)-∫(0,1) x/(1+x^2)dx
=π/4-(1/2)*ln|1+x^2||(0,1)
=π/4-(1/2)*ln2
（3）原式=∫(0,1) x^2d(e^x)
=x^2*e^x|(0,1)-∫(0,1) 2xe^xdx
=e-∫(0,1) 2xd(e^x)
=e-2xe^x|(0,1)+∫(0,1)2e^xdx
=e-2e+2e^x|(0,1)
=-e+2e-2
=e-2
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