What is the integral from 15 to 31 of 1/(sqrt(x-6)) ?

The integral from 15 to 31 of 1/(sqrt(x-6)) is 4

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integral from 15 to 31 of 1/(sqrt(x-6))

\int_{15}^{31} \frac{1}{x - 6} d x

4

Solution steps

\int_{15}^{31} \frac{1}{x - 6} d x

Apply u-substitution

: \int_{3}^{5} 2 d u

= \int_{3}^{5} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{3}^{5}

Compute the boundaries:

4

= 4

\int_{0}^{4} 0.5 e^{- 0.5 x} d x

\int_{- 1}^{1} \frac{2 x}{( x + 3 ) ^{2}} d x

\int_{1}^{2} x^{2} \frac{3}{7} x^{2} d x

\int_{0}^{t} 2 sin (t) d t

\int_{0}^{\frac{π}{2}} 2 sin^{2} (x) d x

What is the integral from 15 to 31 of 1/(sqrt(x-6)) ?
The integral from 15 to 31 of 1/(sqrt(x-6)) is 4