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

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

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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