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

The integral from 0 to 1 of 1/(sqrt(2x+1)) is sqrt(3)-1

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

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

3 - 1

Decimal

0.73205 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} 1 d u

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

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{1}^{3}

Simplify

= [u]_{1}^{3}

Compute the boundaries:

3 - 1

= 3 - 1

\int_{- 1}^{1} sin (2 x) + cos (3 x) d x

\int_{1}^{2} 4 e^{3 x - 2} d x

\int_{3}^{10} 1

\int_{- 1}^{4} 2 x + 4 d x

\int_{1}^{4} \frac{ln ( x )}{x ^{3}} d x

What is the integral from 0 to 1 of 1/(sqrt(2x+1)) ?
The integral from 0 to 1 of 1/(sqrt(2x+1)) is sqrt(3)-1