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

The integral of x/(sqrt((2x-1))) is 1/3 (x+1)sqrt(2x-1)+C

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

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

\frac{1}{3} (x + 1) 2 x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 1}{2} d u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u^{2} d u + \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{1}{2} (\frac{u ^{3}}{3} + u)

Substitute back

u = 2 x - 1

= \frac{1}{2} (\frac{( 2 x - 1 ) ^{3}}{3} + 2 x - 1)

Simplify

\frac{1}{2} (\frac{( 2 x - 1 ) ^{3}}{3} + 2 x - 1) : \frac{1}{3} (x + 1) 2 x - 1

= \frac{1}{3} (x + 1) 2 x - 1

Add a constant to the solution

= \frac{1}{3} (x + 1) 2 x - 1 + C

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