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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 2 d x = 2 x

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

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

Simplify

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{- 2}}{x})

\int \frac{x ^{7} + 2}{( x ^{2} + x + 1 ) ^{2}} d x

\frac{d y}{d x} = \frac{cos ( 20 x )}{e ^{20 y}}

\int t sin (t^{2}) cos (t^{2}) d t

8 - x, x, [0, 4]

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