What is the integral of (3x^6+4x^2+12)/(x^4) ?

The integral of (3x^6+4x^2+12)/(x^4) is x^3-4/x-4/(x^3)+C

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integral of (3x^6+4x^2+12)/(x^4)

\int \frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} d x

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

Solution steps

\int \frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} d x

Expand

\frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} : 3 x^{2} + \frac{4}{x ^{2}} + \frac{12}{x ^{4}}

= \int 3 x^{2} + \frac{4}{x ^{2}} + \frac{12}{x ^{4}} 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 3 x^{2} d x + \int \frac{4}{x ^{2}} d x + \int \frac{12}{x ^{4}} d x

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

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

\int \frac{12}{x ^{4}} d x = - \frac{4}{x ^{3}}

= x^{3} - \frac{4}{x} - \frac{4}{x ^{3}}

Add a constant to the solution

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

What is the integral of (3x^6+4x^2+12)/(x^4) ?
The integral of (3x^6+4x^2+12)/(x^4) is x^3-4/x-4/(x^3)+C