What is the integral of \sqrt[4]{x^3}+1 ?

The integral of \sqrt[4]{x^3}+1 is 4/7 x^{7/4}+x+C

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integral of \sqrt[4]{x^3}+1

\frac{4}{7} x^{\frac{7}{4}} + x + 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 1 d x = x

= \frac{4}{7} x^{\frac{7}{4}} + x

Add a constant to the solution

= \frac{4}{7} x^{\frac{7}{4}} + x + C

x \to 0 - lim (x^{3} cos (\frac{2}{x}))

\frac{d}{d x} (\frac{e ^{x}}{4 x ^{3}})

\int \frac{1}{1 - tan ( x )} d x

\frac{d}{d x} (\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}})

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

What is the integral of \sqrt[4]{x^3}+1 ?
The integral of \sqrt[4]{x^3}+1 is 4/7 x^{7/4}+x+C