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 \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x - 1

= \frac{1}{2} \cdot 2 (2 x - 1)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (2 x - 1)^{\frac{1}{2}} : 2 x - 1

= 2 x - 1

Add a constant to the solution

= 2 x - 1 + C

ln ∣ sin (x) ∣

\int x^{4} (2 x^{5} - 5)^{4} d x

\frac{d}{d x} (\frac{x}{x ^{2} - 1})

\int ln (\frac{2}{x}) d x

\frac{( 2 x ^{3} + 4 x ^{2} - 2 x + 3 )}{( x - 3 )}

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