What is the integral of x/(sqrt(2x-6)) ?

The integral of x/(sqrt(2x-6)) is 1/3 (x+6)sqrt(2x-6)+C

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integral of x/(sqrt(2x-6))

\int \frac{x}{2 x - 6} d x

\frac{1}{3} (x + 6) 2 x - 6 + C

Solution steps

\int \frac{x}{2 x - 6} d x

Apply u-substitution

= \int \frac{u ^{2} + 6}{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} + 6 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 6 d u)

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

\int 6 d u = 6 u

= \frac{1}{2} (\frac{u ^{3}}{3} + 6 u)

Substitute back

u = 2 x - 6

= \frac{1}{2} (\frac{( 2 x - 6 ) ^{3}}{3} + 6 2 x - 6)

Simplify

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

= \frac{1}{3} (x + 6) 2 x - 6

Add a constant to the solution

= \frac{1}{3} (x + 6) 2 x - 6 + C

What is the integral of x/(sqrt(2x-6)) ?
The integral of x/(sqrt(2x-6)) is 1/3 (x+6)sqrt(2x-6)+C