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

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

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integral of 1/(x^3)+x

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\frac{x - 3}{2}

\frac{\partial}{\partial x} (x^{3} y z^{2} + 7 yz)

\frac{d}{d x} (x^{2} e^{x + 3})

y^{^{''}} - 2 y^{^{'}} + y = 1

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{y x}

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