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

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

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

\frac{x ^{3}}{3} + \frac{4}{5} x^{\frac{5}{4}} + 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 x^{2} d x = \frac{x ^{3}}{3}

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

Add a constant to the solution

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

\int (2 t^{2} + 1) d t

\int r \cdot e^{\frac{r}{2}} d r

\int \frac{3 x}{3 x + 5} d x

\int 2 2 x + 3 d x

\int \frac{( x + 1 )}{x ^{2} + 4 x - 5} d x

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