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

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

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integral of x-1-ln(x)

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

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

Solution steps

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

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

\int 1 d x = x

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (y + 3)

y = x^{2} - x, y = 2 (x + 2)

x \to \infty lim (- 4)

\int_{- π}^{π} e^{t} cos (n t) d t

e^{- π s}

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