What is the integral of x*(ln(x))^2 ?

The integral of x*(ln(x))^2 is 1/2 x^2ln^2(x)-1/2 x^2ln(x)+(x^2)/4+C

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integral of x*(ln(x))^2

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

\frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4} + C

Solution steps

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

Apply Integration By Parts

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

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

\int x ln (x) d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}

= \frac{1}{2} x^{2} ln^{2} (x) - (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})

Simplify

= \frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}

Add a constant to the solution

= \frac{1}{2} x^{2} ln^{2} (x) - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4} + C

\frac{d}{d x} (\frac{x + 1}{x ^{2} - y + 1})

y^{^{''}} - 6 y^{^{'}} + 9 = 0

(- 5) (0.7)

\int \frac{1}{e ^{v}} d v

\frac{\partial}{\partial y} (\frac{x ^{2} + y ^{2}}{x + y})

What is the integral of x*(ln(x))^2 ?
The integral of x*(ln(x))^2 is 1/2 x^2ln^2(x)-1/2 x^2ln(x)+(x^2)/4+C