What is the integral of (x^4+2x^3-7x^2+5x-2)/x ?

The integral of (x^4+2x^3-7x^2+5x-2)/x is (x^4)/4+(2x^3)/3-(7x^2)/2+5x-2ln|x|+C

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integral of (x^4+2x^3-7x^2+5x-2)/x

\int \frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} d x

\frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} d x

Expand

\frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} : x^{3} + 2 x^{2} - 7 x + 5 - \frac{2}{x}

= \int x^{3} + 2 x^{2} - 7 x + 5 - \frac{2}{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 2 x^{2} d x - \int 7 x d x + \int 5 d x - \int \frac{2}{x} d x

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

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

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 5 d x = 5 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣ + C

\int \frac{- 8 x ^{2}}{x ^{4} - 1} d x

\int \frac{2}{- x ^{2} - 20 x - 9} d x

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

\int (2 x^{3} - 5 x^{2} + 3 x - 1) d x

\int (7 x^{5} + 2)^{3} d x

What is the integral of (x^4+2x^3-7x^2+5x-2)/x ?
The integral of (x^4+2x^3-7x^2+5x-2)/x is (x^4)/4+(2x^3)/3-(7x^2)/2+5x-2ln|x|+C