ƒ’(x)=arctan(x-1) f(1)=0 求∫f(x)dx 积分区间是（0,1）

-1/4pai-(1/2)ln2

好久没做过反三角函数了，忘记了，哈哈

f(x)=∫f'(x)dx=∫arctan(x-1)dx=xarctan(x-1)-∫x*1/[1+(x-1)^2]dx
=xarctan(x-1)-∫[(x-1)+1]/[1+(x-1)^2]dx=xarctan(x-1)-∫[(x-1)/[1+(x-1)^2]dx-)-∫1/[1+(x-1)^2]dx
=xarctan(x-1)-1/2*ln[1+(x-1)^2]-arctan(x-1)+C=(x-1)arctan(x-1)-1/2*ln[1+(x-1)^2]+C
由f(1)=0得C=0,故f(x)=(x-1)arctan(x-1)-1/2*ln[1+(x-1)^2]
故∫(0,1) f(x)dx=∫(0,1) {(x-1)arctan(x-1)-1/2*ln[1+(x-1)^2]}dx （令t=x-1）
=∫(-1,0) [tarctant-1/2*ln(1+t^2)]dt
∫[tarctant-1/2*ln(1+t^2)]dt=∫arctantd(1/2*t^2)-1/2*t*ln(1+t^2)+∫t*1/(1+t^2)*2tdt
=1/2*t^2*arctant-∫1/2*t^2*1/(1+t^2)dt-1/2*t*ln(1+t^2)+∫2t^2/(1+t^2)dt
=1/2*t^2*arctant-1/2*t*ln(1+t^2)+3/2*∫[1-1/(1+t^2)]dt
=1/2*t^2*arctant-1/2*t*ln(1+t^2)+3t/2-3/2*arctant+C
=(t^2-3)/2*arctant+t/2*[3-ln(1+t^2)]+C
故∫(0,1) f(x)dx=∫(-1,0) [tarctant-1/2*ln(1+t^2)]dt
={(t^2-3)/2*arctant+t/2*[3-ln(1+t^2)]}|(-1,0)
=-π/4

∫f(x)dx =∫{0,1) arctan(x-1)d( x-1)
=∫[-1,0] arctanxdx
=xarctanx-∫x/(1+x^2)^0.5dx
=xarctanx- (1+x^2)^0.5
=2^0.5 -1-Pi/4