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

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

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integral of (1-x)ln(x)

\int (1 - x) ln (x) d x

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

Solution steps

\int (1 - x) ln (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{d}{d x} (2 x sin (x))

\frac{\partial}{\partial y} (sin (x^{3} y^{2}))

\frac{\partial}{\partial y} (3 x^{2} y^{4} - 4)

\frac{2}{x ^{2} + 3}

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