What is the integral from 3 to 4 of 1/(sqrt(x-3)) ?

The integral from 3 to 4 of 1/(sqrt(x-3)) is 2

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integral from 3 to 4 of 1/(sqrt(x-3))

\int_{3}^{4} \frac{1}{x - 3} d x

2

Solution steps

\int_{3}^{4} \frac{1}{x - 3} d x

Apply u-substitution

: \int_{0}^{1} 2 d u

= \int_{0}^{1} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{1}

Compute the boundaries:

2

= 2

\int_{0}^{π} (x - sin (x)) d x

\int_{2}^{3} x^{- 2} d x

\int_{1}^{x^{3}} \frac{1}{1 + t ^{4}} d t

\int_{0}^{1} (- x^{2} + 1) d x

\int_{- 3}^{3} ln (x + 4) d x

What is the integral from 3 to 4 of 1/(sqrt(x-3)) ?
The integral from 3 to 4 of 1/(sqrt(x-3)) is 2