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

The integral from 2 to 8 of x/(x+sqrt(2x)) is 4ln(3/2)+2

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integral from 2 to 8 of x/(x+sqrt(2x))

\int_{2}^{8} \frac{x}{x + 2 x} d x

4 ln (\frac{3}{2}) + 2

Decimal Notation

3.62186 \dots

Solution steps

\int_{2}^{8} \frac{x}{x + 2 x} d x

Apply u-substitution

: \int_{4}^{16} \frac{u}{2 ( u + 2 u )} d u

= \int_{4}^{16} \frac{u}{2 ( u + 2 u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4}^{16} \frac{u}{u + 2 u} d u

Simplify

\frac{u}{u + 2 u} : \frac{u}{u + 2}

= \frac{1}{2} \cdot \int_{4}^{16} \frac{u}{u + 2} d u

Apply u-substitution

: \int_{4}^{6} \frac{2 ( v ^{2} - 4 v + 4 )}{v} d v

= \frac{1}{2} \cdot \int_{4}^{6} \frac{2 ( v ^{2} - 4 v + 4 )}{v} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int_{4}^{6} \frac{v ^{2} - 4 v + 4}{v} d v

Expand

\frac{v ^{2} - 4 v + 4}{v} : v - 4 + \frac{4}{v}

= \frac{1}{2} \cdot 2 \cdot \int_{4}^{6} v - 4 + \frac{4}{v} d v

Apply the Sum Rule:

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

= \frac{1}{2} \cdot 2 (\int_{4}^{6} v d v - \int_{4}^{6} 4 d v + \int_{4}^{6} \frac{4}{v} d v)

\int_{4}^{6} v d v = 10

\int_{4}^{6} 4 d v = 8

\int_{4}^{6} \frac{4}{v} d v = 4 ln (\frac{3}{2})

= \frac{1}{2} \cdot 2 (10 - 8 + 4 ln (\frac{3}{2}))

Simplify

\frac{1}{2} \cdot 2 (10 - 8 + 4 ln (\frac{3}{2})) : 4 ln (\frac{3}{2}) + 2

= 4 ln (\frac{3}{2}) + 2

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What is the integral from 2 to 8 of x/(x+sqrt(2x)) ?
The integral from 2 to 8 of x/(x+sqrt(2x)) is 4ln(3/2)+2