What is the integral from 0 to 4 of 1/(1+sqrt(2x+1)) ?

The integral from 0 to 4 of 1/(1+sqrt(2x+1)) is 2-ln(2)

integral from 0 to 4 of 1/(1+sqrt(2x+1))

\int_{0}^{4} \frac{1}{1 + 2 x + 1} d x

2 - ln (2)

Decimal

1.30685 \dots

Solution steps

\int_{0}^{4} \frac{1}{1 + 2 x + 1} d x

Apply u-substitution

= \int_{1}^{3} \frac{u}{1 + u} d u

Apply u-substitution

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

Expand

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

= \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

= \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)

= 2 - ln (2)

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What is the integral from 0 to 4 of 1/(1+sqrt(2x+1)) ?
The integral from 0 to 4 of 1/(1+sqrt(2x+1)) is 2-ln(2)