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

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

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

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

\frac{5 x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x ∣ - \frac{3}{x} + C

Solution steps

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

Expand

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

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

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{5 x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x ∣ - \frac{3}{x}

Add a constant to the solution

= \frac{5 x ^{3}}{3} + \frac{x ^{2}}{2} - x + ln ∣ x ∣ - \frac{3}{x} + C

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