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

The integral of 1/(sqrt(2x+1)) is sqrt(2x+1)+C

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integral of 1/(sqrt(2x+1))

\int \frac{1}{2 x + 1} d x

2 x + 1 + C

Solution steps

\int \frac{1}{2 x + 1} d x

Apply u-substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = 2 x + 1

= 1 \cdot 2 x + 1

Simplify

= 2 x + 1

Add a constant to the solution

= 2 x + 1 + C

f (x) = \frac{1}{2 x - 5}

\frac{d}{d x} (2 x^{2})

3 x + 5

\frac{d y}{d t} = 4 y (1 - y)

\int_{1}^{2} \frac{2}{x ^{3} + x} d x

What is the integral of 1/(sqrt(2x+1)) ?
The integral of 1/(sqrt(2x+1)) is sqrt(2x+1)+C