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

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

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

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

\frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} - 3 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

= \frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} - 3 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} - 3 ln ∣ x ∣ + C

\int (6 x y + 4 x) d y

(- 5, - 8), (9, - 8)

x \to - 9 lim (5 x - 4)

- \frac{75}{s ^{2} + 6 s + 25}

\frac{d y}{d x} = \frac{( x ^{2} + x y + y ^{2} )}{x ^{2}}

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