What is the integral of 1/(x^5)-3/(x^2) ?

The integral of 1/(x^5)-3/(x^2) is -1/(4x^4)+3/x+C

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integral of 1/(x^5)-3/(x^2)

\int \frac{1}{x ^{5}} - \frac{3}{x ^{2}} d x

- \frac{1}{4 x ^{4}} + \frac{3}{x} + C

Solution steps

\int \frac{1}{x ^{5}} - \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{1}{x ^{5}} d x - \int \frac{3}{x ^{2}} d x

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

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

= - \frac{1}{4 x ^{4}} - (- \frac{3}{x})

Simplify

= - \frac{1}{4 x ^{4}} + \frac{3}{x}

Add a constant to the solution

= - \frac{1}{4 x ^{4}} + \frac{3}{x} + C

x y^{^{'}} + 2 y = x^{2} - x + 1

f (x) = ln (x^{2} - 9 x + 1), (9, 0)

(- 2, 2), (1, - 2)

6 t^{2}

\int \frac{1}{484 - x ^{2}} d x

What is the integral of 1/(x^5)-3/(x^2) ?
The integral of 1/(x^5)-3/(x^2) is -1/(4x^4)+3/x+C