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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - \frac{x ^{4}}{4} - \frac{4}{x}

Add a constant to the solution

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

\int \frac{e ^{4 x}}{e ^{4 x} - 8} d x

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

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

\int (x^{2} - 4 x)^{5} \cdot (4 x - 8) d x

\int \frac{1}{1 + u} d u

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