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

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

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

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

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

Solution steps

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

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

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

= - \frac{x ^{4}}{4} - \frac{1}{x}

Add a constant to the solution

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

\int \frac{4 e ^{- \frac{3}{x ^{5}}}}{x ^{6}} d x

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

3 t + 4

f (x) = x^{3} + x^{2} + x + 1

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