Question

三道定积分。

Answer 1

=∫(-1到1)f(u)d(u+1)
=∫(-1到0)1/(u+2)du+∫(0到1)ln(1+u)du

追答

2.=∫(0到ln2)t/(e^t-2)²de^t
=-∫td1/(e^t-2)
=-t/(e^t-2)+∫1/(e^t-2)dt
=-t/(e^t-2)-∫1/(1-2e^(-t))de^(-t)
=-t/(e^t-2)+(1/2)ln|1-2e^(-t))|

然后定积分为求t=ln2时的极限

7.=∫√4cos²xdx
=∫(0到π/2)2cosxdx+∫(π/2到π)-2cosxdx

=2+2=4