What is the integral from 2 to 23 of 1/(sqrt(x+2)) ?

The integral from 2 to 23 of 1/(sqrt(x+2)) is 6

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integral from 2 to 23 of 1/(sqrt(x+2))

\int_{2}^{23} \frac{1}{x + 2} d x

6

Solution steps

\int_{2}^{23} \frac{1}{x + 2} d x

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

6

= 6

\int_{\frac{1}{2}}^{3} e^{2 x + 3} d x

\int_{- 4}^{- 1} x^{2} (3 - 4 x) d x

\int_{0}^{1} (- x^{3} - 2 x^{2} + 3 x) d x

\int_{1}^{3 x} (x^{3} e^{x y}) d y

\int_{1}^{4} 1 + \frac{1}{u} + \frac{1}{u ^{2}} d u

What is the integral from 2 to 23 of 1/(sqrt(x+2)) ?
The integral from 2 to 23 of 1/(sqrt(x+2)) is 6