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)

\int (4 x^{3} + 1) d x

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

Solution steps

\int 4 x^{3} + 1 d x

Apply the Sum Rule:

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

= \int 4 x^{3} d x + \int 1 d x

\int 4 x^{3} d x = \frac{4}{7} x^{\frac{7}{4}}

\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

\frac{d}{d x} (\frac{1}{( x ^{3} - x + 1 ) ^{2}})^{4}

\int 3 x^{2} (2 x^{2} + \frac{1}{x ^{3}}) d x

\frac{d}{d x} (16 - x^{\frac{2}{3}})

cos ((π x)^{6})

\frac{d}{d x} (x x^{2} - 16)

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