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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 2 d x = 2 x

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{2 x - 2 x} d x

x \to 0 lim (ln (e^{x} - 1))

\frac{1}{( x - 1 )}, at x = 2

x \to 1 + lim (\frac{x}{x ^{2} - 1})

\frac{\partial}{\partial v} ((u^{2} v - v^{3})^{5})

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