What is the integral of 1/(w^{5/8)-w^{1/8}} ?

The integral of 1/(w^{5/8)-w^{1/8}} is

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integral of 1/(w^{5/8)-w^{1/8}}

\int \frac{1}{w ^{\frac{5}{8}} - w ^{\frac{1}{8}}} d w

8 (\frac{1}{3} w^{\frac{3}{8}} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1) + C

Solution steps

\int \frac{1}{w ^{\frac{5}{8}} - w ^{\frac{1}{8}}} d w

Apply u-substitution

= \int \frac{8 u ^{6}}{u ^{4} - 1} d u

Take the constant out:

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

= 8 \cdot \int \frac{u ^{6}}{u ^{4} - 1} d u

Long division

\frac{u ^{6}}{u ^{4} - 1} : u^{2} + \frac{u ^{2}}{u ^{4} - 1}

= 8 \cdot \int u^{2} + \frac{u ^{2}}{u ^{4} - 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

= 8 (\int u^{2} d u + \int \frac{u ^{2}}{u ^{4} - 1} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

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

= 8 (\frac{u ^{3}}{3} + \frac{1}{2} arctan (u) - \frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{4} ln ∣ u - 1 ∣)

Substitute back

u = w^{\frac{1}{8}}

= 8 \frac{( w ^{\frac{1}{8}} ) ^{3}}{3} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1

Simplify

8 \frac{( w ^{\frac{1}{8}} ) ^{3}}{3} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1 : 8 (\frac{1}{3} w^{\frac{3}{8}} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1)

= 8 (\frac{1}{3} w^{\frac{3}{8}} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1)

Add a constant to the solution

= 8 (\frac{1}{3} w^{\frac{3}{8}} + \frac{1}{2} arctan (w^{\frac{1}{8}}) - \frac{1}{4} ln w^{\frac{1}{8}} + 1 + \frac{1}{4} ln w^{\frac{1}{8}} - 1) + C

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What is the integral of 1/(w^{5/8)-w^{1/8}} ?
The integral of 1/(w^{5/8)-w^{1/8}} is