Question

求定积分：∫dx/f(x),上限2,下限1.已知∫f(x)lnxdx=arctanx+c

Answer 1

∫f(x)lnxdx=arctanx+c

等式左右对x求导,则

f(x)lnx=1/(x^2+1)

1/f(x)=lnx(x^2+1)

∫dx/f(x)=∫lnx(x^2+1)dx=lnx[(x^3/3)+x)]-∫[1+1/(3x^2)]dx=lnx[(x^3/3)+x)]-(x^3/9)-x

代入上下限,得

∫dx/f(x)=(14/3)ln2-16/9