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

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

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

t^{\frac{2}{3}} - 3 arctan (t) + 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{3}{1 + t ^{2}} d t = 3 arctan (t)

= t^{\frac{2}{3}} - 3 arctan (t)

Add a constant to the solution

= t^{\frac{2}{3}} - 3 arctan (t) + C

\frac{d}{d x} (200)

x x

\frac{d}{d x} (a e^{- x})

\frac{\partial}{\partial x} (e^{6 y} sin (- 6 x))

n = 1 \sum \infty 17 - \frac{7}{n ^{3}}

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