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

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

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

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

22

Decimal Notation

2.82842 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} 2 d u

= \int_{0}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{2}

Compute the boundaries:

22

= 22

\frac{d y}{d t} - 2 t y = 15 t^{2} e^{t^{2}}

\int x^{3} + x d x

\int \frac{2 x}{x ^{4} + 36} d x

\frac{1}{s} \cdot (e^{- 2 \cdot s \cdot π} - 1)

\frac{d P}{d t} = (P - P^{2})

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