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

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

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

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

22 - 2

Decimal Notation

0.82842 \dots

Solution steps

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

2 9 + x^{2}

\int_{0}^{2 π} sin (x) d x

\int 3 sin (3 x) + 2 cos (4 x) d x

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = sin (3 t)

\frac{3 x}{x}

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