What is the integral of (6x^2+2x-1)/(3x^2) ?

The integral of (6x^2+2x-1)/(3x^2) is 1/3 (6x+2ln|x|+1/x)+C

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integral of (6x^2+2x-1)/(3x^2)

\int \frac{6 x ^{2} + 2 x - 1}{3 x ^{2}} d x

\frac{1}{3} (6 x + 2 ln ∣ x ∣ + \frac{1}{x}) + C

Solution steps

\int \frac{6 x ^{2} + 2 x - 1}{3 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{6 x ^{2} + 2 x - 1}{x ^{2}} d x

Expand

\frac{6 x ^{2} + 2 x - 1}{x ^{2}} : 6 + \frac{2}{x} - \frac{1}{x ^{2}}

= \frac{1}{3} \cdot \int 6 + \frac{2}{x} - \frac{1}{x ^{2}} 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

= \frac{1}{3} (\int 6 d x + \int \frac{2}{x} d x - \int \frac{1}{x ^{2}} d x)

\int 6 d x = 6 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= \frac{1}{3} (6 x + 2 ln ∣ x ∣ - (- \frac{1}{x}))

Simplify

= \frac{1}{3} (6 x + 2 ln ∣ x ∣ + \frac{1}{x})

Add a constant to the solution

= \frac{1}{3} (6 x + 2 ln ∣ x ∣ + \frac{1}{x}) + C

\frac{d}{d x} (\frac{y}{x ^{2} + y ^{2} + 1})

f (x) = x^{2} e^{4 x}

\frac{d}{d x} (ln (x) sec (x))

\frac{d y}{d x} = 3 x^{2} - x y

\int x^{6} cos (x) d x

What is the integral of (6x^2+2x-1)/(3x^2) ?
The integral of (6x^2+2x-1)/(3x^2) is 1/3 (6x+2ln|x|+1/x)+C