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

The integral from-1 to 0 of x/(sqrt(x+1)) is -4/3

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

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

- \frac{4}{3}

Decimal Notation

- 1.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 2 (u^{2} - 1) d u

= \int_{0}^{1} 2 (u^{2} - 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{0}^{1} u^{2} d u - \int_{0}^{1} 1 d u)

\int_{0}^{1} u^{2} d u = \frac{1}{3}

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

= 2 (\frac{1}{3} - 1)

Simplify

2 (\frac{1}{3} - 1) : - \frac{4}{3}

= - \frac{4}{3}

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What is the integral from-1 to 0 of x/(sqrt(x+1)) ?
The integral from-1 to 0 of x/(sqrt(x+1)) is -4/3