Question

x^2除以1+xdx上限1 下限0的定积分

Answer 1

∫(0->1) x^2/(1+x) dx

=∫(0->1) (x^2-1 +1)/(1+x) dx

=∫(0->1) (x-1) dx +∫(0->1) 1/(1+x) dx

=(x^2/2 -x)|(0->1) +ln(x+1)|(0->1)

=(1/2-1) +(ln2-ln1)

=ln2 -1/2