求定积分，0~1，（根号x）*ln（x+1）dx
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满意请采纳，谢谢
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看不清，但是也谢谢啦
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原式=∫(0,1) ln(x+1) d[2/3*x^(3/2)]
=ln(x+1)*(2/3)*x^(3/2)|(0,1)-∫(0,1) (2/3)*x^(3/2)/(x+1) dx
=(2/3)*ln2-(2/3)*∫(0,1) x^(3/2)/(x+1) dx
令x=t^2 dx=2tdt
原式=(2/3)*ln2-(4/3)*∫(0,1) t^4/(t^2+1) dt
=(2/3)*ln2-(4/3)*∫(0,1) (t^4-1+1)/(t^2+1) dt
=(2/3)*ln2-(4/3)*∫(0,1) (t^2-1) dt-(4/3)*∫(0,1) dt/(t^2+1)
=(2/3)*ln2-(4/3)*(t^3/3-t)|(0,1)-(4/3)*arctant|(0,1)
=(2/3)*ln2+8/9-π/3
=ln(x+1)*(2/3)*x^(3/2)|(0,1)-∫(0,1) (2/3)*x^(3/2)/(x+1) dx
=(2/3)*ln2-(2/3)*∫(0,1) x^(3/2)/(x+1) dx
令x=t^2 dx=2tdt
原式=(2/3)*ln2-(4/3)*∫(0,1) t^4/(t^2+1) dt
=(2/3)*ln2-(4/3)*∫(0,1) (t^4-1+1)/(t^2+1) dt
=(2/3)*ln2-(4/3)*∫(0,1) (t^2-1) dt-(4/3)*∫(0,1) dt/(t^2+1)
=(2/3)*ln2-(4/3)*(t^3/3-t)|(0,1)-(4/3)*arctant|(0,1)
=(2/3)*ln2+8/9-π/3
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