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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

2 t e^{3 t} + 4 t^{2} e^{- 5 t}

\frac{d}{d x} (\frac{1}{2 x - 3})

5 - 4 x

e^{x^{5}}

\int sin (x^{2}) d x

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