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

The integral of 1/2 x^2ln(x)-(x^2)/4-(x^2)/2 is 1/2 (1/3 x^3ln(x)-(x^3)/9)-(x^3}{12}-\frac{x^3)/6+C

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integral of 1/2 x^2ln(x)-(x^2)/4-(x^2)/2

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

\frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}) - \frac{x ^{3}}{12} - \frac{x ^{3}}{6} + C

Solution steps

\int \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2} 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{x ^{2}}{4} d x - \int \frac{x ^{2}}{2} 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{x ^{2}}{4} d x = \frac{x ^{3}}{12}

\int \frac{x ^{2}}{2} d x = \frac{x ^{3}}{6}

= \frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}) - \frac{x ^{3}}{12} - \frac{x ^{3}}{6}

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}) - \frac{x ^{3}}{12} - \frac{x ^{3}}{6} + C

\int cos (x) sin (x) d x

x \to - 5 lim (\frac{∣ x + 5 ∣}{x + 5})

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 5

\int 1 + \frac{1}{x} d x

What is the integral of 1/2 x^2ln(x)-(x^2)/4-(x^2)/2 ?
The integral of 1/2 x^2ln(x)-(x^2)/4-(x^2)/2 is 1/2 (1/3 x^3ln(x)-(x^3)/9)-(x^3}{12}-\frac{x^3)/6+C