What is the integral of 1/(sqrt(y-1)) ?

The integral of 1/(sqrt(y-1)) is 2sqrt(y-1)+C

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integral of 1/(sqrt(y-1))

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

2 y - 1 + C

Solution steps

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

Apply u-substitution

: \int 2 d u

= \int 2 d u

Integral of a constant:

\int a d x = a x

= 2 u

Substitute back

u = y - 1

= 2 y - 1

Add a constant to the solution

= 2 y - 1 + C

x \to 1 lim (1^{2} - 3)

x y^{^{'}} - 2 y = - x^{2}

\int e^{- x^{5}} d x

\int \frac{1}{t ^{2} 16 - t ^{2}} d t

y = 2 - x^{3}, (1, 1)

What is the integral of 1/(sqrt(y-1)) ?
The integral of 1/(sqrt(y-1)) is 2sqrt(y-1)+C