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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d u} (x^{2} + x y (x, u) (x, u)^{3})

\int t ln (t) d t

\int x^{\frac{5}{4}} d x

t^{2} \frac{d ^{2} y}{d t ^{2}} + 2 t \frac{d y}{d t} - 6 y = 0

t an g e n t (x + 2)^{2} + (y - 3)^{2} = 37, (- 8, 4)

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