What is the integral of 1/(sqrt(x-16)) ?

The integral of 1/(sqrt(x-16)) is 2sqrt(x-16)+C

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integral of 1/(sqrt(x-16))

\int \frac{1}{x - 16} d x

2 x - 16 + C

Solution steps

\int \frac{1}{x - 16} d x

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 = x - 16

= 2 x - 16

Add a constant to the solution

= 2 x - 16 + C

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What is the integral of 1/(sqrt(x-16)) ?
The integral of 1/(sqrt(x-16)) is 2sqrt(x-16)+C