若f'(X)=ln(x^2+1),且f(0)=0,则积分上限1下限0f(X)dx=

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/04 06:28:15
若f'(X)=ln(x^2+1),且f(0)=0,则积分上限1下限0f(X)dx=
xݓ]kPǿJ)L%MMoD99T EQ+ldND+{P+ uJ/yu?焔^Unjwp^gUjh Dsyvz<+U["7.JB+6.-:Mf8؁tcjBnnG:\g"1rDqEӴEӶ8y)y9n$,) 3i sMH Q :1l'C)Ew%`Qx}}snb8'zJ_  w/o@l~gD1r_;\{+

若f'(X)=ln(x^2+1),且f(0)=0,则积分上限1下限0f(X)dx=
若f'(X)=ln(x^2+1),且f(0)=0,则积分上限1下限0f(X)dx=

若f'(X)=ln(x^2+1),且f(0)=0,则积分上限1下限0f(X)dx=
f'(x)=ln(x^2+1)
f(x)=∫ln(x^2+1)dx=xln(x^2+1)-∫2x^2/(x^2+1)dx
=xln(x^2+1)-∫(2x^2+2-2)/(x^2+1)dx=xln(x^2+1)-2x+2arctanx+C
f(0)=0代入得:C=0
∫(0,1)f(x)dx=∫(0,1)(xln(x^2+1)-2x+2arctanx)dx
=[(1/2)x^2ln(x^2+1)-(1/2)ln(x^2+1)-3x^2/2+2xarctanx]|(0,1)
=3/2-π/2

 过程大同小异  就是计算结果算出:2/π-2/5     

收起