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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

= \frac{3 x ^{4}}{4} - x^{2} - \frac{3}{x}

Add a constant to the solution

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

\frac{d}{d x} (4^{x} - 7)

\frac{d y}{d x} + \frac{y}{x} = 3 x^{9} y^{2}

\int_{- 4}^{4} (1 + 16 - x^{2}) d x

\frac{d}{d x} (\frac{8 x}{( x ^{2} + 4 ) ^{2}})

\int \frac{x ^{5}}{x ^{6} + 7} d x

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