What is the integral of (sqrt(x)-2x^2+1)/x ?

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

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integral of (sqrt(x)-2x^2+1)/x

\int \frac{x - 2 x ^{2} + 1}{x} d x

2 x - x^{2} + ln ∣ x ∣ + C

Solution steps

\int \frac{x - 2 x ^{2} + 1}{x} d x

Expand

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= 2 x - x^{2} + ln ∣ x ∣

Add a constant to the solution

= 2 x - x^{2} + ln ∣ x ∣ + C

r \to \infty lim (ln (r))

\int \frac{( x - 1 )}{( x ^{3} - x ^{2} - 2 x )} d x

f (x) = (x^{3} - 7^{x} + 3), at x = 1

y^{^{''}} - 9 y^{^{'}} = 7 t

f (x) = \frac{tan ( x ^{2} )}{4}

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