What is the integral from 1 to 6 of (2x)/(sqrt(x+3)) ?

The integral from 1 to 6 of (2x)/(sqrt(x+3)) is 40/3

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integral from 1 to 6 of (2x)/(sqrt(x+3))

\int_{1}^{6} \frac{2 x}{x + 3} d x

\frac{40}{3}

Decimal

13.33333 \dots

Solution steps

\int_{1}^{6} \frac{2 x}{x + 3} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{6} \frac{x}{x + 3} d x

Apply u-substitution

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

\int_{2}^{3} u^{2} d u = \frac{19}{3}

\int_{2}^{3} 3 d u = 3

= 2 \cdot 2 (\frac{19}{3} - 3)

Simplify

2 \cdot 2 (\frac{19}{3} - 3) : \frac{40}{3}

= \frac{40}{3}

What is the integral from 1 to 6 of (2x)/(sqrt(x+3)) ?
The integral from 1 to 6 of (2x)/(sqrt(x+3)) is 40/3