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

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

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

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

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

Solution steps

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

Expand

\frac{3 x - 2}{x ^{5}} : \frac{3}{x ^{4}} - \frac{2}{x ^{5}}

= \int \frac{3}{x ^{4}} - \frac{2}{x ^{5}} 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 ^{5}} d x

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

\int \frac{2}{x ^{5}} d x = - \frac{1}{2 x ^{4}}

= - \frac{1}{x ^{3}} - (- \frac{1}{2 x ^{4}})

Simplify

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

Add a constant to the solution

= - \frac{1}{x ^{3}} + \frac{1}{2 x ^{4}} + C

\frac{d}{d x} (2 cos (x) - x sin (x))

n = 0 \sum \infty (\frac{6}{7})^{n}

\frac{d y}{d x} + 12 y = 4

(2 x y^{2} + 2 y) d x + (2 x^{2} y + 2 x) d y = 0, y (1) = 2

y^{^{''}} - 5 y^{^{'}} + 6 y = - sin (5 t)

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