h(x)= integral from 0 to e^x of ln(t) dt

solution

let u = e^x then du/dx =e^x

also dh/dx=dh/du du/dx

h'(x)=d/dx integral from 1 to e^x of ln(t) dt = d/du integral from 1 to u ln(t) dt times du/dx = ln u du/dx = (lne^x)*e^x = xe^x

for "d/du integral from 1 to u ln(t) dt times du/dx" above, why is the integral multiplied by the derivative of u WRT x