What is the integral of (t^2+\sqrt[4]{t}) ?

The integral of (t^2+\sqrt[4]{t}) is (t^3)/3+4/5 t^{5/4}+C

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integral of (t^2+\sqrt[4]{t})

\int (t^{2} + 4 t) d t

\frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}} + C

Solution steps

\int t^{2} + 4 t d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{2} d t + \int 4 t d t

\int t^{2} d t = \frac{t ^{3}}{3}

\int 4 t d t = \frac{4}{5} t^{\frac{5}{4}}

= \frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}}

Add a constant to the solution

= \frac{t ^{3}}{3} + \frac{4}{5} t^{\frac{5}{4}} + C

n = 1 \sum \infty \frac{1}{2 ^{n}}

\frac{\partial}{\partial y} (n sin (2 (m x + n y)))

\frac{30}{( 10 s + 1 ) ( s ^{2} + 4 )}

x^{x + 2}

\frac{d}{d x} (\frac{x ^{2} - 64}{x - 8})

What is the integral of (t^2+\sqrt[4]{t}) ?
The integral of (t^2+\sqrt[4]{t}) is (t^3)/3+4/5 t^{5/4}+C