What is the integral of 1/(x^{4/5)(1+x^{1/5})} ?

The integral of 1/(x^{4/5)(1+x^{1/5})} is 5ln|x^{1/5}+1|+C

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integral of 1/(x^{4/5)(1+x^{1/5})}

\int \frac{1}{x ^{\frac{4}{5}} ( 1 + x ^{\frac{1}{5}} )} d x

5 ln x^{\frac{1}{5}} + 1 + C

Solution steps

\int \frac{1}{x ^{\frac{4}{5}} ( 1 + x ^{\frac{1}{5}} )} d x

Apply u-substitution

: \int \frac{5}{4 u ^{\frac{3}{4}} ( u ^{\frac{1}{4}} + 1 )} d u

= \int \frac{5}{4 u ^{\frac{3}{4}} ( u ^{\frac{1}{4}} + 1 )} d u

Take the constant out:

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

= \frac{5}{4} \cdot \int \frac{1}{u ^{\frac{3}{4}} ( u ^{\frac{1}{4}} + 1 )} d u

Apply u-substitution

: \int \frac{4}{v} d v

= \frac{5}{4} \cdot \int \frac{4}{v} d v

Take the constant out:

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

= \frac{5}{4} \cdot 4 \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{5}{4} \cdot 4 ln ∣ v ∣

Substitute back

= \frac{5}{4} \cdot 4 ln (x^{\frac{4}{5}})^{\frac{1}{4}} + 1

Simplify

\frac{5}{4} \cdot 4 ln (x^{\frac{4}{5}})^{\frac{1}{4}} + 1 : 5 ln x^{\frac{1}{5}} + 1

= 5 ln x^{\frac{1}{5}} + 1

Add a constant to the solution

= 5 ln x^{\frac{1}{5}} + 1 + C

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What is the integral of 1/(x^{4/5)(1+x^{1/5})} ?
The integral of 1/(x^{4/5)(1+x^{1/5})} is 5ln|x^{1/5}+1|+C