What is the integral of 1/(x^{3/4)-x^{1/4}} ?

The integral of 1/(x^{3/4)-x^{1/4}} is -4(1/2 ln|x^{1/4}+1|-1/2 ln|x^{1/4}-1|-x^{1/4})+C

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integral of 1/(x^{3/4)-x^{1/4}}

\int \frac{1}{x ^{\frac{3}{4}} - x ^{\frac{1}{4}}} d x

- 4 (\frac{1}{2} ln x^{\frac{1}{4}} + 1 - \frac{1}{2} ln x^{\frac{1}{4}} - 1 - x^{\frac{1}{4}}) + C

Solution steps

\int \frac{1}{x ^{\frac{3}{4}} - x ^{\frac{1}{4}}} d x

Apply u-substitution

: \int \frac{4 u ^{2}}{u ^{2} - 1} d u

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

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= 4 \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 4 (- \int \frac{u ^{2}}{- u ^{2} + 1} d u)

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= 4 (- \int \frac{1}{- u ^{2} + 1} - 1 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

= 4 (- (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

= 4 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

Substitute back

u = x^{\frac{1}{4}}

= 4 - \frac{ln x ^{\frac{1}{4}} + 1}{2} - \frac{ln x ^{\frac{1}{4}} - 1}{2} - x^{\frac{1}{4}}

Simplify

4 - \frac{ln x ^{\frac{1}{4}} + 1}{2} - \frac{ln x ^{\frac{1}{4}} - 1}{2} - x^{\frac{1}{4}} : - 4 (\frac{1}{2} ln x^{\frac{1}{4}} + 1 - \frac{1}{2} ln x^{\frac{1}{4}} - 1 - x^{\frac{1}{4}})

= - 4 (\frac{1}{2} ln x^{\frac{1}{4}} + 1 - \frac{1}{2} ln x^{\frac{1}{4}} - 1 - x^{\frac{1}{4}})

Add a constant to the solution

= - 4 (\frac{1}{2} ln x^{\frac{1}{4}} + 1 - \frac{1}{2} ln x^{\frac{1}{4}} - 1 - x^{\frac{1}{4}}) + C

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What is the integral of 1/(x^{3/4)-x^{1/4}} ?
The integral of 1/(x^{3/4)-x^{1/4}} is -4(1/2 ln|x^{1/4}+1|-1/2 ln|x^{1/4}-1|-x^{1/4})+C