What is the integral of (x^5-x^3+6x)/(x^4) ?

The integral of (x^5-x^3+6x)/(x^4) is (x^2)/2-ln|x|-3/(x^2)+C

integral of (x^5-x^3+6x)/(x^4)

\int \frac{x ^{5} - x ^{3} + 6 x}{x ^{4}} d x

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

Solution steps

\int \frac{x ^{5} - x ^{3} + 6 x}{x ^{4}} d x

Expand

\frac{x ^{5} - x ^{3} + 6 x}{x ^{4}} : x - \frac{1}{x} + \frac{6}{x ^{3}}

= \int x - \frac{1}{x} + \frac{6}{x ^{3}} 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 d x - \int \frac{1}{x} d x + \int \frac{6}{x ^{3}} d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x ^{3}} d x = - \frac{3}{x ^{2}}

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

Add a constant to the solution

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

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

\frac{d}{d x} (2 x + \frac{1}{x ^{2}})

\int_{2}^{\infty} e^{- 8 p} d p

n = 1 \sum \infty (\frac{5}{9})^{n}

\int - \frac{1}{2} ln ∣ x ∣ - 8 x + 2 x^{2} + C^{1} d x

What is the integral of (x^5-x^3+6x)/(x^4) ?
The integral of (x^5-x^3+6x)/(x^4) is (x^2)/2-ln|x|-3/(x^2)+C