What is the integral of (6x^4+4x^2-2)/(x^2) ?

The integral of (6x^4+4x^2-2)/(x^2) is 2x^3+4x+2/x+C

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integral of (6x^4+4x^2-2)/(x^2)

\int \frac{6 x ^{4} + 4 x ^{2} - 2}{x ^{2}} d x

2 x^{3} + 4 x + \frac{2}{x} + C

Solution steps

\int \frac{6 x ^{4} + 4 x ^{2} - 2}{x ^{2}} d x

Expand

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

= \int 6 x^{2} + 4 - \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

= \int 6 x^{2} d x + \int 4 d x - \int \frac{2}{x ^{2}} d x

\int 6 x^{2} d x = 2 x^{3}

\int 4 d x = 4 x

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

= 2 x^{3} + 4 x - (- \frac{2}{x})

Simplify

= 2 x^{3} + 4 x + \frac{2}{x}

Add a constant to the solution

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

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

\int (x \cdot e^{2 x}) d x

\int \frac{1}{( x + 3 ) ^{4}} d x

\int x (cos (x) + 1) d x

\int a^{2} x d x

What is the integral of (6x^4+4x^2-2)/(x^2) ?
The integral of (6x^4+4x^2-2)/(x^2) is 2x^3+4x+2/x+C