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

The integral of 1/(sqrt(x)\sqrt{x-1)} is 2ln|sqrt(x)+sqrt(x-1)|+C

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

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

2 ln x + x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 1} d u

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

Take the constant out:

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

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

Use the common integral:

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

= 2 ln u^{2} + 1 + u

Substitute back

u = x - 1

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

(x - 1)^{2} + 1 = x

= 2 ln x + x - 1

Add a constant to the solution

= 2 ln x + x - 1 + C

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