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_{1}^{2} x - 1 d x

\frac{3 s}{( s ^{2} + 4 )}

y^{^{''''}} + 2 y^{^{''}} + y = 0

\frac{d}{d x} ((\frac{10}{4 + x})^{\frac{1}{2}})

- \frac{1}{4} x^{4} + 6 x^{2}

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