What is the integral of ((x^5+2x^2-1))/(x^4) ?

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

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integral of ((x^5+2x^2-1))/(x^4)

\int \frac{( x ^{5} + 2 x ^{2} - 1 )}{x ^{4}} d x

\frac{x ^{2}}{2} - \frac{2}{x} + \frac{1}{3 x ^{3}} + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

2 y^{^{'}} + y = \frac{3}{2} t^{2}

x \to - 2 lim (\frac{2 x}{x + 2})

\frac{d}{d t} (cos (4 t))

\frac{d}{d x} (11 x^{\frac{1}{2}} + x^{\frac{3}{2}})

\int_{- 6}^{- 2} (- x^{2} - 8 x - 18) d x

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