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

The integral of 3/(x^{3/4)}-4\sqrt[3]{x}+1 is 12\sqrt[4]{x}-3x^{4/3}+x+C

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integral of 3/(x^{3/4)}-4\sqrt[3]{x}+1

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 \frac{3}{x ^{\frac{3}{4}}} d x = 12 x^{\frac{1}{4}}

\int 1 d x = x

= 12 x^{\frac{1}{4}} - 3 x^{\frac{4}{3}} + x

Apply radical rule:

Add a constant to the solution

f (x) = \frac{4 x - 3}{5 x ^{2} + 3 x}

\int x sin^{3} (x) d x

\int x^{2} arctan (x) d x

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

\frac{d ^{3}}{d x ^{3}} (\frac{7}{x ^{2}})

What is the integral of 3/(x^{3/4)}-4\sqrt[3]{x}+1 ?
The integral of 3/(x^{3/4)}-4\sqrt[3]{x}+1 is 12\sqrt[4]{x}-3x^{4/3}+x+C