What is the integral of (12x^3+8x^2-6)/(x^2) ?

The integral of (12x^3+8x^2-6)/(x^2) is 6x^2+8x+6/x+C

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integral of (12x^3+8x^2-6)/(x^2)

\int \frac{12 x ^{3} + 8 x ^{2} - 6}{x ^{2}} d x

6 x^{2} + 8 x + \frac{6}{x} + C

Solution steps

\int \frac{12 x ^{3} + 8 x ^{2} - 6}{x ^{2}} d x

Expand

\frac{12 x ^{3} + 8 x ^{2} - 6}{x ^{2}} : 12 x + 8 - \frac{6}{x ^{2}}

= \int 12 x + 8 - \frac{6}{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 12 x d x + \int 8 d x - \int \frac{6}{x ^{2}} d x

\int 12 x d x = 6 x^{2}

\int 8 d x = 8 x

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

= 6 x^{2} + 8 x - (- \frac{6}{x})

Simplify

= 6 x^{2} + 8 x + \frac{6}{x}

Add a constant to the solution

= 6 x^{2} + 8 x + \frac{6}{x} + C

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What is the integral of (12x^3+8x^2-6)/(x^2) ?
The integral of (12x^3+8x^2-6)/(x^2) is 6x^2+8x+6/x+C