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

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

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

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

x^{2} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= x^{2} + ln ∣ x ∣

Add a constant to the solution

= x^{2} + ln ∣ x ∣ + C

x \to 1 lim (\frac{x ^{4} - 1}{x ^{2}})

f (x) = (2 x^{2} - 4)^{- 1}

x \to 8 - lim (\frac{x}{x ^{2} - 64})

\frac{d}{d x} (\frac{- 2 x ^{2} + 9}{9 - x ^{2}})

y^{^{'}} = (y + 2 t)^{2}, y (0) = 3