What is the integral of 1/(sqrt(\sqrt{x)-1)} ?

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

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integral of 1/(sqrt(\sqrt{x)-1)}

\int \frac{1}{x - 1} d x

4 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1) + C

Solution steps

\int \frac{1}{x - 1} d x

Apply u-substitution

: \int 4 (u^{2} + 1) d u

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

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

\int 1 d u = u

= 4 (\frac{u ^{3}}{3} + u)

Substitute back

u = x - 1

= 4 \frac{( x - 1 ) ^{3}}{3} + x - 1

Simplify

4 \frac{( x - 1 ) ^{3}}{3} + x - 1 : 4 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1)

= 4 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1)

Add a constant to the solution

= 4 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1) + C

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