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

The integral of (2x^4+6x^3-x)/(x^3) is x^2+6x+1/x+C

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

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

x^{2} + 6 x + \frac{1}{x} + C

Solution steps

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

Expand

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

= \int 2 x + 6 - \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

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

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

\int 6 d x = 6 x

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

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

Simplify

= x^{2} + 6 x + \frac{1}{x}

Add a constant to the solution

= x^{2} + 6 x + \frac{1}{x} + C

\frac{d}{d x} (2 x + \frac{2}{x})

y = e^{x}, y = e^{2 x - 1}, x = 0

\int cos (\frac{1}{2}) d x

\int (\frac{3}{2 x - 1}) d x

\frac{d y}{d x}, x^{2} + y^{2} = 2 x y

What is the integral of (2x^4+6x^3-x)/(x^3) ?
The integral of (2x^4+6x^3-x)/(x^3) is x^2+6x+1/x+C