What is the integral of (3x^4-x^3+6x^2)/(x^4) ?

The integral of (3x^4-x^3+6x^2)/(x^4) is 3x-ln|x|-6/x+C

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integral of (3x^4-x^3+6x^2)/(x^4)

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

3 x - ln ∣ x ∣ - \frac{6}{x} + C

Solution steps

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

Expand

\frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} : 3 - \frac{1}{x} + \frac{6}{x ^{2}}

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

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

\int 3 d x = 3 x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= 3 x - ln ∣ x ∣ - \frac{6}{x}

Add a constant to the solution

= 3 x - ln ∣ x ∣ - \frac{6}{x} + C

y = 12 - x^{2}, y = x^{2} - 6

\frac{d}{d x} ((cos (7 x))^{x})

x, \frac{1}{2} x, 0, 25

\frac{\partial}{\partial x} (\frac{3 x - y}{3 x + y})

\int x (2 x - 1)^{9} d x

What is the integral of (3x^4-x^3+6x^2)/(x^4) ?
The integral of (3x^4-x^3+6x^2)/(x^4) is 3x-ln|x|-6/x+C