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

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

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

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

2

Solution steps

\int_{0}^{4} \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:

2

= 2

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\int_{0}^{2} e^{- y} d y

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

\int_{1}^{3} 5 x^{4} d x

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

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