What is the integral from 0 to 1 of 1/(sqrt(x+2)) ?

The integral from 0 to 1 of 1/(sqrt(x+2)) is 2sqrt(3)-2sqrt(2)

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integral from 0 to 1 of 1/(sqrt(x+2))

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

23 - 22

Decimal

0.63567 \dots

Solution steps

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

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

23 - 22

= 23 - 22

\frac{7 s + 4}{2 s ^{2} + 16 s + 30}

\frac{x - 2}{x ^{2} - 4}

x \to 2 lim (- x + 2 x^{2} - 1)

\frac{10}{s ^{2} + 2 s + 10}

\frac{d}{d x} ((\frac{1}{x ^{2}})^{t a n (x)})

What is the integral from 0 to 1 of 1/(sqrt(x+2)) ?
The integral from 0 to 1 of 1/(sqrt(x+2)) is 2sqrt(3)-2sqrt(2)