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

The integral of (x^4-5x^2+2x)/(5x^2) is (x^3)/(15)-x+2/5 ln|x|+C

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

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

\frac{x ^{3}}{15} - x + \frac{2}{5} ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

= \frac{1}{5} \cdot \int x^{2} - 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

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

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

\int 5 d x = 5 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{1}{5} (\frac{x ^{3}}{3} - 5 x + 2 ln ∣ x ∣)

Simplify

\frac{1}{5} (\frac{x ^{3}}{3} - 5 x + 2 ln ∣ x ∣) : \frac{x ^{3}}{15} - x + \frac{2}{5} ln ∣ x ∣

= \frac{x ^{3}}{15} - x + \frac{2}{5} ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{3}}{15} - x + \frac{2}{5} ln ∣ x ∣ + C

6 + \frac{4}{x} + \frac{6}{x ^{2}}

\int 4^{x} (4^{x} + 1)^{3} d x

(- \frac{1}{2}, 1), (- 2, \frac{1}{2})

\frac{d}{d x} (sin (x^{3} + 2))

\frac{\partial}{\partial x} (4 x - 2)

What is the integral of (x^4-5x^2+2x)/(5x^2) ?
The integral of (x^4-5x^2+2x)/(5x^2) is (x^3)/(15)-x+2/5 ln|x|+C