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

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

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

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

\frac{x ^{2}}{2} + 5 x + \frac{4}{x} + C

Solution steps

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

Expand

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

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

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

\int 5 d x = 5 x

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

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

Simplify

= \frac{x ^{2}}{2} + 5 x + \frac{4}{x}

Add a constant to the solution

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

\int 30 x + sin (x) + C d x

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

x \to π lim (6 cot (x))

\frac{d ^{2} y}{d x ^{2}} = 1 + \frac{d y}{d x}

x \to 0 lim (\frac{1}{sin ^{2} ( 2 x )})

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