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

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

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

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

2 x - x^{2} + C

Solution steps

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

Expand

\frac{x - 2 x ^{2}}{x} : \frac{1}{x} - 2 x

= \int \frac{1}{x} - 2 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 \frac{1}{x} d x - \int 2 x d x

\int \frac{1}{x} d x = 2 x

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

= 2 x - x^{2}

Add a constant to the solution

= 2 x - x^{2} + C

\int arccos (y) d y

\int_{- 2}^{3} 2 x^{2} + 5 d x

x \to - 7 lim (\frac{1}{x + 7})

\frac{d}{d x} (cot^{3} (π - x))

x \to - 3 lim (\frac{1}{x - 1})

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