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

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

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

\int (3 - \frac{1}{x ^{4}} + \frac{1}{x ^{2}}) d x

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

Solution steps

\int 3 - \frac{1}{x ^{4}} + \frac{1}{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 3 d x - \int \frac{1}{x ^{4}} d x + \int \frac{1}{x ^{2}} d x

\int 3 d x = 3 x

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

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

Simplify

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

Add a constant to the solution

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

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