1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
问题描述：
1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
答
∫(1/x^2)cos(1/x)dx=-∫cos(1/x)d(1/x)=-sin(1/x)+C2.∵(arcsinx)'=xf(x)=(1-x^2)^ (-1/2)∴f(x)=[x (1-x^2)^ 1/2] ^(-1)1/f(x)=x(1-x^2) ^1/2∫1/f(x)dx =∫x(1-x^2) ^1/2dx=-1/2∫(1-x^2)^ 1/2 d(1-x^2)= -1/3(1-...