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

\frac{d y}{d x} = (x - 2) e^{- 2 y}

\frac{\partial}{\partial x} ((2 - x) (5 - x - 2 y))

x \to 2 lim (\frac{3 x + 4}{5 x + 7})

\frac{d}{d x} (\frac{1}{cos ( x )} + \frac{1}{sin ( x + 1 )})

\frac{d}{d θ} (\frac{e ^{i θ} + e ^{- i θ}}{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