What is the integral of (2x)/(sqrt(1-6x)) ?

The integral of (2x)/(sqrt(1-6x)) is 1/27 ((-6x+1)^{3/2}-3sqrt(-6x+1))+C

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integral of (2x)/(sqrt(1-6x))

\int \frac{2 x}{1 - 6 x} d x

\frac{1}{27} ((- 6 x + 1)^{\frac{3}{2}} - 3 - 6 x + 1) + C

Solution steps

\int \frac{2 x}{1 - 6 x} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{1 - 6 x} d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{2} - 1}{18} d u

Take the constant out:

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

= 2 \cdot \frac{1}{18} \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

= 2 \cdot \frac{1}{18} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= 2 \cdot \frac{1}{18} (\frac{u ^{3}}{3} - u)

Substitute back

u = 1 - 6 x

= 2 \cdot \frac{1}{18} (\frac{( 1 - 6 x ) ^{3}}{3} - 1 - 6 x)

Simplify

2 \cdot \frac{1}{18} (\frac{( 1 - 6 x ) ^{3}}{3} - 1 - 6 x) : \frac{1}{27} ((- 6 x + 1)^{\frac{3}{2}} - 3 - 6 x + 1)

= \frac{1}{27} ((- 6 x + 1)^{\frac{3}{2}} - 3 - 6 x + 1)

Add a constant to the solution

= \frac{1}{27} ((- 6 x + 1)^{\frac{3}{2}} - 3 - 6 x + 1) + C

What is the integral of (2x)/(sqrt(1-6x)) ?
The integral of (2x)/(sqrt(1-6x)) is 1/27 ((-6x+1)^{3/2}-3sqrt(-6x+1))+C