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

The integral from 0 to 2 of (2x)/(sqrt(x+1)) is 8/3

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

\int_{0}^{2} \frac{2 x}{x + 1} d x

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

\int_{0}^{2} \frac{2 x}{x + 1} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} \frac{x}{x + 1} d x

Apply u-substitution

: \int_{1}^{3} 2 (u^{2} - 1) d u

= 2 \cdot \int_{1}^{3} 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 2 \cdot \int_{1}^{3} 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 \cdot 2 (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 1 d u)

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

\int_{1}^{3} 1 d u = 3 - 1

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

Simplify

2 \cdot 2 (3 - \frac{1}{3} - (3 - 1)) : \frac{8}{3}

= \frac{8}{3}

\int_{0}^{π} \frac{1}{2 x + π} d x

\int_{1}^{y} (2 x y + y^{2}) d x

\int_{- 1}^{2} x^{3} + 8 d x

\int_{- π}^{π} ∣ t ∣ d t

\int_{0}^{y} x^{2} + y^{2} d x

What is the integral from 0 to 2 of (2x)/(sqrt(x+1)) ?
The integral from 0 to 2 of (2x)/(sqrt(x+1)) is 8/3