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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

\int x + 4 y d x

n = 1 \sum \infty \frac{n ^{n}}{n !}

4 \frac{d y}{d x} + 28 y = 7

\int 4 cos^{3} (2 x) d x

(x^{l n (x)})^{^{'}}

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