Integrate the function f(x) = x ln (x) over the interval [1,e].

This problem can be solved using integration by parts (IBP), although using IBP multiple times:BASIC IBP METHOD: given an expression integral(u dv) where u and v are functions of a common variable x and 'dv' (or u') denotes the 1st derivative of the function v we use the following formula:"integral(u dv) = uv - integral(v du)"we evaluate 'du' (1st derivative of u, also denoted u'), where we know uwe evaluate 'v', where we know 'dv' by integrating vthe method rests upon the expression du being mathematically easy to work with as compared with u (e.g. du = 1), and rests upon dv being easy to integrate STEP ZERO: write down problemI = integral( x ln(x)) over [1,e]STEP ONE: find integral(ln x)Trick: split integral into 1* ln(x)choosing u = ln(x), dv = 1, we have du = 1/x and v = x + const.J = integral(u dv) = uv - integral(v du) = xln(x) - integral (1) = xln(x) - x + constantSTEP TWO: solve the problem using integration by parts Trick: we will find the integral that we want evaluated on both sides after applying the IBP formula. I.e. I = (some terms) - I + (some other terms), which can simplify by writing as 2I = (some terms) + (some other terms).choosing u = x, dv = ln(x), we have du = 1 and v = xln(x) - x + const. (given step one)I = integral(u dv) = uv - integral(v du) = [x²(ln(x) - x)]^e₁ - integral(xln(x) - x)I = [x²(ln(x) - 1)]^e₁ - I + integral(x) (split right-hand integral and apply trick!)2I = e²(1 - 1) - 1(0 - 1) + [0.5 x²]^e₁ (evaluate left hand bracketed expression above and integrate x)I = 0 + 0.5 + 0.25 e² - 0.25 (divide all by 2, evaluate right hand bracketed expression, simplify the overall expression)I = 0.25 e² + 0.25 (simplify terms with same powers of e)STEP THREE:check working!