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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n !}{n \cdot 3 ^{n}}

y = 18 sin (x), y = 18 cos (x)

x^{2} - 4 x + 3, - x^{2} + 2 x + 3, 4, 0

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 8

y = \frac{ln ( 6 x )}{6 x}

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