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

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

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

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

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

Solution steps

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

Expand

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\int (4 - 4 cos (x))^{2} d x

y = 4 x - x^{2}, at x = a

\frac{d}{d x} (2 x (4 x + 9)^{\frac{1}{2}} + 12)

\frac{d}{d x} ((x^{2} + x) \cdot (tan (x)))

\frac{d}{d x} (arcsin (- x))

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