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

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

y^{^{''}} - 4 y^{^{'}} + 7 y = x e^{x}

f (x) = \frac{1}{x ^{31}}

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

x \to 4 lim (3 x^{3} - 2 x^{2} + x - 7)

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