What is the integral of x(\sqrt[4]{x^3}+x^{-2}) ?

The integral of x(\sqrt[4]{x^3}+x^{-2}) is 1/3 (12/11 x^{11/4}+ln|x^3|)+C

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integral of x(\sqrt[4]{x^3}+x^{-2})

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

\frac{1}{3} (\frac{12}{11} x^{\frac{11}{4}} + ln x^{3}) + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{\frac{2}{3}} 4 u + 1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{u ^{\frac{2}{3}} 4 u + 1}{u} d u

Expand

\frac{u ^{\frac{2}{3}} 4 u + 1}{u} : \frac{1}{12 u} + \frac{1}{u}

= \frac{1}{3} \cdot \int \frac{1}{12 u} + \frac{1}{u} d u

Apply the Sum Rule:

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

= \frac{1}{3} (\int \frac{1}{12 u} d u + \int \frac{1}{u} d u)

\int \frac{1}{12 u} d u = \frac{12}{11} u^{\frac{11}{12}}

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{3} (\frac{12}{11} u^{\frac{11}{12}} + ln ∣ u ∣)

Substitute back

u = x^{3}

= \frac{1}{3} (\frac{12}{11} (x^{3})^{\frac{11}{12}} + ln x^{3})

(x^{3})^{\frac{11}{12}} = x^{\frac{11}{4}}

= \frac{1}{3} (\frac{12}{11} x^{\frac{11}{4}} + ln x^{3})

Add a constant to the solution

= \frac{1}{3} (\frac{12}{11} x^{\frac{11}{4}} + ln x^{3}) + C

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What is the integral of x(\sqrt[4]{x^3}+x^{-2}) ?
The integral of x(\sqrt[4]{x^3}+x^{-2}) is 1/3 (12/11 x^{11/4}+ln|x^3|)+C