What is the integral of 1/(sqrt(h-19)) ?

The integral of 1/(sqrt(h-19)) is 2sqrt(h-19)+C

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integral of 1/(sqrt(h-19))

\int \frac{1}{h - 19} d h

2 h - 19 + C

Solution steps

\int \frac{1}{h - 19} d h

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 = h - 19

= 2 h - 19

Add a constant to the solution

= 2 h - 19 + C

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What is the integral of 1/(sqrt(h-19)) ?
The integral of 1/(sqrt(h-19)) is 2sqrt(h-19)+C