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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (4 z)

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

f^{^{'}} (x) = x^{\frac{2}{3}}

y = x^{4} + 4 e^{x}, (0, 4)

\int \frac{x + 3}{x ^{2} + 6 x} d x

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