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

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

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

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

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

Solution steps

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

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

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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

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