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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

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

\frac{\partial}{\partial u} (u^{2} - v^{2})

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

\frac{1}{1 + t}

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