What is the integral of (4x^5+2x^3+x-1)/(x^2) ?

The integral of (4x^5+2x^3+x-1)/(x^2) is x^4+x^2+ln|x|+1/x+C

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integral of (4x^5+2x^3+x-1)/(x^2)

\int \frac{4 x ^{5} + 2 x ^{3} + x - 1}{x ^{2}} d x

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

Solution steps

\int \frac{4 x ^{5} + 2 x ^{3} + x - 1}{x ^{2}} d x

Expand

\frac{4 x ^{5} + 2 x ^{3} + x - 1}{x ^{2}} : 4 x^{3} + 2 x + \frac{1}{x} - \frac{1}{x ^{2}}

= \int 4 x^{3} + 2 x + \frac{1}{x} - \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 4 x^{3} d x + \int 2 x d x + \int \frac{1}{x} d x - \int \frac{1}{x ^{2}} d x

\int 4 x^{3} d x = x^{4}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

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

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

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

\frac{d}{d x} ((2 x - 5)^{\frac{1}{2}})

\int y^{2} e^{- 2 y} d y

What is the integral of (4x^5+2x^3+x-1)/(x^2) ?
The integral of (4x^5+2x^3+x-1)/(x^2) is x^4+x^2+ln|x|+1/x+C