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

The integral of u^{-2}(2u^4+3u^2-2u^{-3}) is -(2u^7+3u^5-2}{u^4}+\frac{8u^3)/3+6u-3/(2u^4)+C

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

\int u^{- 2} (2 u^{4} + 3 u^{2} - 2 u^{- 3}) d u

- \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} + \frac{8 u ^{3}}{3} + 6 u - \frac{3}{2 u ^{4}} + C

Solution steps

\int u^{- 2} (2 u^{4} + 3 u^{2} - 2 u^{- 3}) d u

Apply Integration By Parts

: - \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} - \int - \frac{8 u ^{7} + 6 u ^{5} + 6}{u ^{5}} d u

= - \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} - \int - \frac{8 u ^{7} + 6 u ^{5} + 6}{u ^{5}} d u

\int - \frac{8 u ^{7} + 6 u ^{5} + 6}{u ^{5}} d u = - \frac{8 u ^{3}}{3} - 6 u + \frac{3}{2 u ^{4}}

= - \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} - (- \frac{8 u ^{3}}{3} - 6 u + \frac{3}{2 u ^{4}})

Simplify

= - \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} + \frac{8 u ^{3}}{3} + 6 u - \frac{3}{2 u ^{4}}

Add a constant to the solution

= - \frac{2 u ^{7} + 3 u ^{5} - 2}{u ^{4}} + \frac{8 u ^{3}}{3} + 6 u - \frac{3}{2 u ^{4}} + C

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

\frac{d}{d x} (x^{2} e^{- 2 x})

x \to \frac{1}{2} lim (∣ 2 x + 3 ∣)

\frac{\partial}{\partial x} (x^{6} y^{8} + 7 x^{8} y)

- \frac{y}{x}

What is the integral of u^{-2}(2u^4+3u^2-2u^{-3}) ?
The integral of u^{-2}(2u^4+3u^2-2u^{-3}) is -(2u^7+3u^5-2}{u^4}+\frac{8u^3)/3+6u-3/(2u^4)+C