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

The integral from 3 to 8 of x/(sqrt(1+x)) is 32/3

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integral from 3 to 8 of x/(sqrt(1+x))

\int_{3}^{8} \frac{x}{1 + x} d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{3}^{8} \frac{x}{1 + x} d x

Apply u-substitution

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

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

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

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

= 2 (\frac{19}{3} - 1)

Simplify

2 (\frac{19}{3} - 1) : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 3 to 8 of x/(sqrt(1+x)) ?
The integral from 3 to 8 of x/(sqrt(1+x)) is 32/3