What is the integral from 1 to 16 of 1/(3-sqrt(x)) ?

The integral from 1 to 16 of 1/(3-sqrt(x)) is diverges

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integral from 1 to 16 of 1/(3-sqrt(x))

\int_{1}^{16} \frac{1}{3 - x} d x

diverges

Solution steps

\int_{1}^{16} \frac{1}{3 - x} d x

Find undefined (singularity) points:

x = 9

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{1}^{9} \frac{1}{3 - x} d x + \int_{9}^{16} \frac{1}{3 - x} d x

\int_{1}^{9} \frac{1}{3 - x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

4 t \cdot \frac{d y}{d t} - y = t

x \to 0 lim (tan (5 x))

x \to 1 - lim (3 x - 1)

y = x^{2} - x^{3}, (0, 1)

\frac{\partial}{\partial x} (sec^{2} (x))

What is the integral from 1 to 16 of 1/(3-sqrt(x)) ?
The integral from 1 to 16 of 1/(3-sqrt(x)) is diverges