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

The integral of ((x^5-x^3+2x))/(x^4) is (x^2)/2-ln|x|-1/(x^2)+C

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{x ^{2}}{2} - ln ∣ x ∣ - \frac{1}{x ^{2}}

Add a constant to the solution

= \frac{x ^{2}}{2} - ln ∣ x ∣ - \frac{1}{x ^{2}} + C

1 - e^{- t}

\int \frac{6 x ^{3} + 8 x}{x ^{4} - 16} d x

\frac{d y}{d x} = 6 x (y - 1)^{\frac{2}{3}}

\int \frac{x ^{3} - 6 x - 20}{x + 5} d x

t sin (π) t

What is the integral of ((x^5-x^3+2x))/(x^4) ?
The integral of ((x^5-x^3+2x))/(x^4) is (x^2)/2-ln|x|-1/(x^2)+C