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

x \to - \infty lim (9 x^{3})

\int \frac{p + 1}{p ^{2} + 1} d p

\frac{d}{d x} (\frac{m}{2 π ( a - x )})

\int_{0}^{2} \frac{7 t}{( t - 3 ) ^{2}} d t

\frac{d}{d x} (- 3 sin (x) - x cos (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