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

The integral from 2 to 8 of 1/(sqrt(2x)+1) is 2-ln(5)+ln(3)

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integral from 2 to 8 of 1/(sqrt(2x)+1)

\int_{2}^{8} \frac{1}{2 x + 1} d x

2 - ln (5) + ln (3)

Decimal Notation

1.48917 \dots

Solution steps

\int_{2}^{8} \frac{1}{2 x + 1} d x

Apply u-substitution

: \int_{4}^{16} \frac{1}{2 ( u + 1 )} d u

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

Apply u-substitution

: \int_{3}^{5} \frac{2 ( v - 1 )}{v} d v

= \frac{1}{2} \cdot \int_{3}^{5} \frac{2 ( v - 1 )}{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_{3}^{5} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= \frac{1}{2} \cdot 2 \cdot \int_{3}^{5} 1 - \frac{1}{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_{3}^{5} 1 d v - \int_{3}^{5} \frac{1}{v} d v)

\int_{3}^{5} 1 d v = 2

\int_{3}^{5} \frac{1}{v} d v = ln (5) - ln (3)

= \frac{1}{2} \cdot 2 (2 - (ln (5) - ln (3)))

Simplify

\frac{1}{2} \cdot 2 (2 - (ln (5) - ln (3))) : 2 - ln (5) + ln (3)

= 2 - ln (5) + ln (3)

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What is the integral from 2 to 8 of 1/(sqrt(2x)+1) ?
The integral from 2 to 8 of 1/(sqrt(2x)+1) is 2-ln(5)+ln(3)