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

The integral from 0 to 1 of sqrt(2/(x+1)) is 4-2sqrt(2)

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

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

4 - 22

Decimal

1.17157 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{2}{x + 1} x - \int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x]_{0}^{1}

= [\frac{2}{x + 1} x - \int - \frac{x}{2 ( x + 1 ) ^{\frac{3}{2}}} d x]_{0}^{1}

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

= [\frac{2}{x + 1} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1}

Simplify

[\frac{2}{x + 1} x - (- \frac{2 ( x + 2 )}{x + 1})]_{0}^{1} : [22 1 + x]_{0}^{1}

= [22 1 + x]_{0}^{1}

Compute the boundaries:

4 - 22

= 4 - 22

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What is the integral from 0 to 1 of sqrt(2/(x+1)) ?
The integral from 0 to 1 of sqrt(2/(x+1)) is 4-2sqrt(2)