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

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

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

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

2

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 2 d u

= \int_{0}^{1} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{1}

Compute the boundaries:

2

= 2

\frac{\partial}{\partial x} (x^{9})

y = 4 x^{3}, (- 1, - 4)

x \to \frac{1}{2} lim (\frac{x + 1}{x + 5})

\frac{d y}{d x} = - \frac{3 y}{x} - 2 - x^{- 4}

\frac{1}{1 - x}, 3

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