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

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

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

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

6 x - 3 ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{6 x ^{2} - 3 x}{x ^{2}} : 6 - \frac{3}{x}

= \int 6 - \frac{3}{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 6 d x - \int \frac{3}{x} d x

\int 6 d x = 6 x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

= 6 x - 3 ln ∣ x ∣

Add a constant to the solution

= 6 x - 3 ln ∣ x ∣ + C

y^{^{'}} = 2 t

f (x) = x + 11

\int - 4 e^{2 x} d x

\frac{d y}{d x} = e^{(y - x)}

\int_{0}^{\infty} x a e^{- a x} d x

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