What is the integral of 1/((yc-1)^{1/2)} ?

The integral of 1/((yc-1)^{1/2)} is (2(cy-1)^{1/2})/c+C

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integral of 1/((yc-1)^{1/2)}

\int \frac{1}{( yc - 1 ) ^{\frac{1}{2}}} d y

\frac{2 ( cy - 1 ) ^{\frac{1}{2}}}{c} + C

Solution steps

\int \frac{1}{( yc - 1 ) ^{\frac{1}{2}}} d y

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = yc - 1

= \frac{1}{c} \cdot 2 (yc - 1)^{\frac{1}{2}}

Simplify

\frac{1}{c} \cdot 2 (yc - 1)^{\frac{1}{2}} : \frac{2 ( cy - 1 ) ^{\frac{1}{2}}}{c}

= \frac{2 ( cy - 1 ) ^{\frac{1}{2}}}{c}

Add a constant to the solution

= \frac{2 ( cy - 1 ) ^{\frac{1}{2}}}{c} + C

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What is the integral of 1/((yc-1)^{1/2)} ?
The integral of 1/((yc-1)^{1/2)} is (2(cy-1)^{1/2})/c+C