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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 x z + y^{2})

\int_{1}^{16} \frac{1}{9 - x} d x

f (x) = tan (3 x), at x = \frac{π}{9}

\int 2 + 2 cos (4 x) d x

\frac{\partial}{\partial y} (\frac{x}{( x ^{2} + y ^{2} )})

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