What is the integral from 5 to 9 of 1/(sqrt(x-5)) ?

The integral from 5 to 9 of 1/(sqrt(x-5)) is 4

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integral from 5 to 9 of 1/(sqrt(x-5))

\int_{5}^{9} \frac{1}{x - 5} d x

4

Solution steps

\int_{5}^{9} \frac{1}{x - 5} d x

Apply u-substitution

: \int_{0}^{2} 2 d u

= \int_{0}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{2}

Compute the boundaries:

4

= 4

\int_{0}^{4} (x^{2} - 4) d x

\int_{2}^{\infty} \frac{1}{ln ( n )}

\int_{0}^{\frac{π}{4}} \frac{1}{x ^{2} + 1} d x

\int_{7}^{9} x^{2} + 2 x - 6 d x

\int_{0}^{a} 2 x a^{2} - x^{2} d x

What is the integral from 5 to 9 of 1/(sqrt(x-5)) ?
The integral from 5 to 9 of 1/(sqrt(x-5)) is 4