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

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

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integral from 1 to 5 of 1/(1+sqrt(2x-1))

\int_{1}^{5} \frac{1}{1 + 2 x - 1} d x

2 - ln (2)

Decimal

1.30685 \dots

Solution steps

\int_{1}^{5} \frac{1}{1 + 2 x - 1} d x

Apply u-substitution

: \int_{1}^{9} \frac{1}{2 ( 1 + u )} d u

= \int_{1}^{9} \frac{1}{2 ( 1 + 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_{1}^{9} \frac{1}{1 + u} d u

Apply u-substitution

: \int_{2}^{4} \frac{2 ( v - 1 )}{v} d v

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

Expand

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

= \frac{1}{2} \cdot 2 \cdot \int_{2}^{4} 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_{2}^{4} 1 d v - \int_{2}^{4} \frac{1}{v} d v)

\int_{2}^{4} 1 d v = 2

\int_{2}^{4} \frac{1}{v} d v = ln (2)

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

Simplify

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

= 2 - ln (2)

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