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

The integral of (3/(x^4)+2-3/(x^2)) is -1/(x^3)+2x+3/x+C

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

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

- \frac{1}{x ^{3}} + 2 x + \frac{3}{x} + C

Solution steps

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

\int \frac{3}{x ^{4}} d x = - \frac{1}{x ^{3}}

\int 2 d x = 2 x

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

= - \frac{1}{x ^{3}} + 2 x - (- \frac{3}{x})

Simplify

= - \frac{1}{x ^{3}} + 2 x + \frac{3}{x}

Add a constant to the solution

= - \frac{1}{x ^{3}} + 2 x + \frac{3}{x} + C

What is the integral of (3/(x^4)+2-3/(x^2)) ?
The integral of (3/(x^4)+2-3/(x^2)) is -1/(x^3)+2x+3/x+C