∫lnx×f(x)dx=arctanx+c 求∫从1到e的1/f(x)dx?注：积分的下限是1，上限是e
热心网友
对∫lnx×f(x)dx=arctanx+c两边求导，则f(x)lnx = 1/(1+x^2)f(x) = 1/(1+x^2)lnx1/f(x) = (1+x^2)lnx∫从1到e的1/f(x)dx= ∫lnxdx + ∫x^2 * lnx dx (1 -- e)= xlnx - ∫dx + (1/3)*∫lnxd(x^3) (1 -- e)= xlnx - x + 1/3[x^3 * lnx - ∫x^2dx] (1 -- e)= xlnx - x + 1/3[x^3 * lnx - x^3/3] (1 -- e)= (2e^3 + 10)/9