求定积分∫01xarctanxdx
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【答案】：解：
∫xarctanxdx=∫arctanxd(x²/2)
=(x²/2)arctanx-∫x²/2darctanx
=(x²/2)arctanx-(1/2)∫x²/(1+x²)dx
=(x²/2)arctanx-(1/2)∫[1-1/(1+x²)]dx
=(x²/2)arctanx-(1/2)(x-arctanx)
=(1/2)(x²arctanx+arctanx-x)|(0~1)
=(1/2)(π/4+π/4-1)
=π/4-1/2
∫xarctanxdx=∫arctanxd(x²/2)
=(x²/2)arctanx-∫x²/2darctanx
=(x²/2)arctanx-(1/2)∫x²/(1+x²)dx
=(x²/2)arctanx-(1/2)∫[1-1/(1+x²)]dx
=(x²/2)arctanx-(1/2)(x-arctanx)
=(1/2)(x²arctanx+arctanx-x)|(0~1)
=(1/2)(π/4+π/4-1)
=π/4-1/2
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