What is the integral of x/(sqrt(4x-2)) ?

The integral of x/(sqrt(4x-2)) is 1/6 (x+1)sqrt(4x-2)+C

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integral of x/(sqrt(4x-2))

\int \frac{x}{4 x - 2} d x

\frac{1}{6} (x + 1) 4 x - 2 + C

Solution steps

\int \frac{x}{4 x - 2} d x

Apply u-substitution

: \int \frac{u ^{2} + 2}{8} d u

= \int \frac{u ^{2} + 2}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{2} + 2 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}{8} (\int u^{2} d u + \int 2 d u)

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

\int 2 d u = 2 u

= \frac{1}{8} (\frac{u ^{3}}{3} + 2 u)

Substitute back

u = 4 x - 2

= \frac{1}{8} (\frac{( 4 x - 2 ) ^{3}}{3} + 2 4 x - 2)

Simplify

\frac{1}{8} (\frac{( 4 x - 2 ) ^{3}}{3} + 2 4 x - 2) : \frac{1}{6} (x + 1) 4 x - 2

= \frac{1}{6} (x + 1) 4 x - 2

Add a constant to the solution

= \frac{1}{6} (x + 1) 4 x - 2 + C

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What is the integral of x/(sqrt(4x-2)) ?
The integral of x/(sqrt(4x-2)) is 1/6 (x+1)sqrt(4x-2)+C