The integral of y = x*ln can be found using integrations by parts which states that: Int[ u dv] = u*v - Int [ v du]

Int[ x*ln x dx]

let u = ln x => du = (1/x)dx

dv = x dx => v = x^2/2

=> ln x * x^2/2 - Int [ (x^2 / 2)(1/x) dx]

=> ln x * x^2/2 - Int [ (x / 2) dx]

=> ln x * x^2/2 - x^2 / 4

=> [2*x^2*ln x - x^2]/4 + C

**The integral of x*ln x = [2*x^2*ln x - x^2]/4 + C**

For solving the integral, we'll apply integration by parts:

Int f(x)dx = Int x*ln xdx

We'll apply the formula of integration by parts:

Int udv = uv - Int vdu

u = ln x

We'll differentiate both sides:

du = dx/x

dv = xdx => Int xdx = Int dv = v

v = x^2/2

We'll substitute u,v,du,dv in the formula above:

Int x*ln xdx = [(x^2)*(ln x)]/2 - Int x^2dx/2x

We'll simplify and we'll get:

Int x*ln xdx = [(x^2)*(ln x)]/2 - x^2/4 + C

We'll factorize by x^2/2:

**Int x*ln xdx = (x^2/2)*(ln x - 1/2) + C**

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