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

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

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

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

22 - 2

Decimal Notation

0.82842 \dots

Solution steps

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

Apply u-substitution

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

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

Integral of a constant:

\int a d x = a x

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

Compute the boundaries:

22 - 2

= 22 - 2

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

\int \frac{( 6 - x )}{4 x ^{2} - 12 x + 7} d x

\int \frac{ln ( 2 x )}{2 x ^{2}} d x

\int t (arctan (t)) d t

\frac{d}{d p} (3 p (1 - 0.5 p))

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