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

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

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

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

\frac{5 x ^{2}}{4} - \frac{3}{2} x - \frac{1}{x} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{5 x ^{3} - 3 x ^{2} + 2}{x ^{2}} d x

Expand

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

= \frac{1}{2} \cdot \int 5 x - 3 + \frac{2}{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

= \frac{1}{2} (\int 5 x d x - \int 3 d x + \int \frac{2}{x ^{2}} d x)

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 3 d x = 3 x

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

= \frac{1}{2} (\frac{5 x ^{2}}{2} - 3 x - \frac{2}{x})

Simplify

\frac{1}{2} (\frac{5 x ^{2}}{2} - 3 x - \frac{2}{x}) : \frac{5 x ^{2}}{4} - \frac{3}{2} x - \frac{1}{x}

= \frac{5 x ^{2}}{4} - \frac{3}{2} x - \frac{1}{x}

Add a constant to the solution

= \frac{5 x ^{2}}{4} - \frac{3}{2} x - \frac{1}{x} + C

\int_{0}^{3} 1 + x^{2} d x

x \to 6 + lim (6 x - 6)

y = \frac{x}{4} - \frac{4}{x}

\frac{( 2 )}{s ^{2} + 5 s + 3}

\frac{d}{d x} (a x + 4)

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