What is the integral of 1/(t^3)+\sqrt[3]{t} ?

The integral of 1/(t^3)+\sqrt[3]{t} is -1/(2t^2)+3/4 t^{4/3}+C

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integral of 1/(t^3)+\sqrt[3]{t}

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

Solution steps

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}{t ^{3}} d t = - \frac{1}{2 t ^{2}}

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

Add a constant to the solution

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

x \to 1 lim (e^{4 x^{3} - 4 x})

\frac{d}{d x} (((x + x^{9})^{- 6} + 3 x)^{3})

\frac{d}{d z} (x z)

4 y y^{^{'}} = x

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

What is the integral of 1/(t^3)+\sqrt[3]{t} ?
The integral of 1/(t^3)+\sqrt[3]{t} is -1/(2t^2)+3/4 t^{4/3}+C