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

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

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

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

x^{2} + 4 x + \frac{1}{x} + C

Solution steps

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

Expand

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

= \int 2 x + 4 - \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 4 d x - \int \frac{1}{x ^{2}} d x

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

\int 4 d x = 4 x

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

= x^{2} + 4 x - (- \frac{1}{x})

Simplify

= x^{2} + 4 x + \frac{1}{x}

Add a constant to the solution

= x^{2} + 4 x + \frac{1}{x} + C

\int 6^{x} e^{x} d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{17}{2}}} d x

y^{^{'}} + y = - 4 cos (x)

x \to \infty lim (1^{x})

(7 x^{6} + 4 x^{3})^{4}

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