Question

ln(1+x^2)dx定积分范围0到1

分部积分法求定积分
求定积分∫ln(1+x^2)dx,积分区间 (0,1)
求定积分∫arctan跟xdx,积分区间 (0,1)
arctan跟号下xdx

Answer 1

1,xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2)-2∫(1-1/(1+x^2))dx
=xln(1+x^2)-2(x-arctanx)
2,设t=√x,x=t^2,dx=2tdt
∫arctan√xdx
=∫2tarctantdt
=∫arctantd(t^2)
=t^2arctant-∫t^2/(1+t^2)dt
=t^2arctant-∫(1-1/(1+t^2)dt
=t^2arctant-t+arctant
=xarctan√x-√x+arctan√x