What is the integral of (\sqrt[4]{x}+1)^2 ?

The integral of (\sqrt[4]{x}+1)^2 is 2/3 x^{3/2}+8/5 x^{5/4}+x+C

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integral of (\sqrt[4]{x}+1)^2

\frac{2}{3} x^{\frac{3}{2}} + \frac{8}{5} x^{\frac{5}{4}} + x + C

Solution steps

Expand

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 d x = \frac{2}{3} x^{\frac{3}{2}}

\int 1 d x = x

= \frac{2}{3} x^{\frac{3}{2}} + \frac{8}{5} x^{\frac{5}{4}} + x

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + \frac{8}{5} x^{\frac{5}{4}} + x + C

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What is the integral of (\sqrt[4]{x}+1)^2 ?
The integral of (\sqrt[4]{x}+1)^2 is 2/3 x^{3/2}+8/5 x^{5/4}+x+C