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

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

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

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

x^{2} + 3 x + \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

\int 3 d x = 3 x

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

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

Simplify

= x^{2} + 3 x + \frac{1}{x}

Add a constant to the solution

= x^{2} + 3 x + \frac{1}{x} + C

f (x) = x + 7, at x = - 6

\frac{d}{d x} (4^{- \frac{x}{4}})

\frac{4 x ^{2} - 3}{8 x ^{3} + 1}

\frac{d}{d x} (\frac{x cos ( π x ) + 1}{1 - x})

\frac{s + 1}{s ^{2} + 2}

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