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

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

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integral of 1/(2sqrt(y-1))

\int \frac{1}{2 y - 1} d y

y - 1 + C

Solution steps

\int \frac{1}{2 y - 1} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{y - 1} d y

Apply u-substitution

: \int 2 d u

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

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 2 u

Substitute back

u = y - 1

= \frac{1}{2} \cdot 2 y - 1

Simplify

\frac{1}{2} \cdot 2 y - 1 : y - 1

= y - 1

Add a constant to the solution

= y - 1 + C

n = 0 \sum \infty 0. 9^{n}

\int tan^{7} (x) sec^{2} (x) d x

\int_{1}^{e} \frac{( x ^{2} + 1 )}{x} d x

x \to 4 + lim (ln (x^{2} - 16))

\frac{1}{2} y^{^{'}} + y = 0, y (1) = 1

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