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

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

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

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

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

Solution steps

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

Expand

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

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

\int 2 ln (x) d x = 2 (x ln (x) - x)

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

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

Add a constant to the solution

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

\frac{d}{d x} (10 sin (x) + 103 cos (x))

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\frac{d}{d x} (ln (4 + x))

\int x e^{x} - e^{x} d x

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