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

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

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

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

\frac{2 x ^{3}}{3} + 3 x + \frac{1}{2 x ^{4}} + C

Solution steps

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

Simplify

\frac{2 x ^{4} + 3 x ^{2} - 2 x ^{- 3}}{x ^{2}} : \frac{2 x ^{7} + 3 x ^{5} - 2}{x ^{5}}

= \int \frac{2 x ^{7} + 3 x ^{5} - 2}{x ^{5}} d x

Expand

\frac{2 x ^{7} + 3 x ^{5} - 2}{x ^{5}} : 2 x^{2} + 3 - \frac{2}{x ^{5}}

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

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

\int 3 d x = 3 x

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

= \frac{2 x ^{3}}{3} + 3 x - (- \frac{1}{2 x ^{4}})

Simplify

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

Add a constant to the solution

= \frac{2 x ^{3}}{3} + 3 x + \frac{1}{2 x ^{4}} + C

n = 1 \sum \infty \frac{1}{6 n + 5}

y^{^{'}} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

\int e^{3 x} sin (3 x) d x

y^{^{'}} = \frac{x}{y}, y (0) = 3

x \to - 2 lim (3 x^{2} + 7 - \frac{1}{x})

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