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∫xarctanxdx
=(1/2)∫ arctanxd(x²)
=(1/2)x²arctanx - (1/2)∫ x²/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ (x²+1-1)/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x²) dx
=(1/2)x²arctanx - (1/2)x + (1/2)arctanx + C
所以定积分为
∫(0,1)xarctanxdx
=(1/2)x²arctanx|(0,1)-(1/2)x|(0,1)+(1/2)arctanx|(0,1)
=(1/2× π/4-0)-(1/2-0)+(1/2× π/4-0)
=π/4-1/2
=(1/2)∫ arctanxd(x²)
=(1/2)x²arctanx - (1/2)∫ x²/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ (x²+1-1)/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x²) dx
=(1/2)x²arctanx - (1/2)x + (1/2)arctanx + C
所以定积分为
∫(0,1)xarctanxdx
=(1/2)x²arctanx|(0,1)-(1/2)x|(0,1)+(1/2)arctanx|(0,1)
=(1/2× π/4-0)-(1/2-0)+(1/2× π/4-0)
=π/4-1/2
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