What is the integral of 1/2 x^2ln(x)-1/2 x^2ln(2x) ?

The integral of 1/2 x^2ln(x)-1/2 x^2ln(2x) is 1/6 x^3ln(x)-1/6 x^3ln(2x)+C

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integral of 1/2 x^2ln(x)-1/2 x^2ln(2x)

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

\frac{1}{6} x^{3} ln (x) - \frac{1}{6} x^{3} ln (2 x) + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2} x^{2} ln (x) d x = \frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9})

\int \frac{1}{2} x^{2} ln (2 x) d x = \frac{1}{2} (\frac{1}{3} x^{3} ln (2 x) - \frac{x ^{3}}{9})

= \frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}) - \frac{1}{2} (\frac{1}{3} x^{3} ln (2 x) - \frac{x ^{3}}{9})

Simplify

\frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}) - \frac{1}{2} (\frac{1}{3} x^{3} ln (2 x) - \frac{x ^{3}}{9}) : \frac{1}{6} x^{3} ln (x) - \frac{1}{6} x^{3} ln (2 x)

= \frac{1}{6} x^{3} ln (x) - \frac{1}{6} x^{3} ln (2 x)

Add a constant to the solution

= \frac{1}{6} x^{3} ln (x) - \frac{1}{6} x^{3} ln (2 x) + C

\int ln (- x) d x

x \to 0 + lim (ln (x) x)

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

\frac{d}{d x} (sec (x) - 1)

\int \frac{2}{x ^{2} + 2} d x

What is the integral of 1/2 x^2ln(x)-1/2 x^2ln(2x) ?
The integral of 1/2 x^2ln(x)-1/2 x^2ln(2x) is 1/6 x^3ln(x)-1/6 x^3ln(2x)+C