What is the integral of x^2log_{e}(x) ?

The integral of x^2log_{e}(x) is 1/3 x^3ln(x)-(x^3)/9+C

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integral of x^2log_{e}(x)

\int x^{2} lo g_{e} (x) d x

\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + C

Solution steps

\int x^{2} lo g_{e} (x) d x

Simplify

lo g_{e} (x) : ln (x)

= \int x^{2} ln (x) d x

Apply Integration By Parts

: \frac{1}{3} x^{3} ln (x) - \int \frac{x ^{2}}{3} d x

= \frac{1}{3} x^{3} ln (x) - \int \frac{x ^{2}}{3} d x

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}

Add a constant to the solution

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + C

\frac{d y}{d t} = y + 1

\int_{0}^{7} 49 - x^{2} d x

\frac{\partial}{\partial x} ((x - 2) (1 - e^{- x}))

\frac{1}{4} y^{^{''}} + 4 t = 2 tan (4 t) - \frac{1}{3} e^{t}

x \to 0 lim (\frac{sec ( x ) - 1}{x})

What is the integral of x^2log_{e}(x) ?
The integral of x^2log_{e}(x) is 1/3 x^3ln(x)-(x^3)/9+C