What is the integral of (2x^5+3x^4-5x^2+8)/(x^2) ?

The integral of (2x^5+3x^4-5x^2+8)/(x^2) is (x^4)/2+x^3-5x-8/x+C

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integral of (2x^5+3x^4-5x^2+8)/(x^2)

\int \frac{2 x ^{5} + 3 x ^{4} - 5 x ^{2} + 8}{x ^{2}} d x

\frac{x ^{4}}{2} + x^{3} - 5 x - \frac{8}{x} + C

Solution steps

\int \frac{2 x ^{5} + 3 x ^{4} - 5 x ^{2} + 8}{x ^{2}} d x

Expand

\frac{2 x ^{5} + 3 x ^{4} - 5 x ^{2} + 8}{x ^{2}} : 2 x^{3} + 3 x^{2} - 5 + \frac{8}{x ^{2}}

= \int 2 x^{3} + 3 x^{2} - 5 + \frac{8}{x ^{2}} 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^{3} d x + \int 3 x^{2} d x - \int 5 d x + \int \frac{8}{x ^{2}} d x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

\int 5 d x = 5 x

\int \frac{8}{x ^{2}} d x = - \frac{8}{x}

= \frac{x ^{4}}{2} + x^{3} - 5 x - \frac{8}{x}

Add a constant to the solution

= \frac{x ^{4}}{2} + x^{3} - 5 x - \frac{8}{x} + C

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\int e^{- 2 x} cos (5 x) d x

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What is the integral of (2x^5+3x^4-5x^2+8)/(x^2) ?
The integral of (2x^5+3x^4-5x^2+8)/(x^2) is (x^4)/2+x^3-5x-8/x+C