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

The integral of (2-2x)/(1-sqrt(x)) is 2x+4/3 x^{3/2}+4+C

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

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

2 x + \frac{4}{3} x^{\frac{3}{2}} + 4 + C

Solution steps

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

Expand

\frac{2 - 2 x}{1 - x} : \frac{2}{1 - x} - \frac{2 x}{1 - x}

Apply the Sum Rule:

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

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

\int \frac{2}{1 - x} d x = - 4 (- 1 + x + ln 1 - x)

\int \frac{2 x}{1 - x} d x = 4 (- \frac{1}{3} x^{\frac{3}{2}} - \frac{x}{2} - x - ln - x + 1)

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

Simplify

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

= 2 x + \frac{4}{3} x^{\frac{3}{2}} + 4

Add a constant to the solution

= 2 x + \frac{4}{3} x^{\frac{3}{2}} + 4 + C

\frac{d y}{d x} + \frac{y}{x} = 2 x^{8} y^{2}

\frac{\partial}{\partial x} (arctan (9 x))

y = (4 x^{2} + 3) (2 x - 3)

\int 18 arctan (x) d x

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

What is the integral of (2-2x)/(1-sqrt(x)) ?
The integral of (2-2x)/(1-sqrt(x)) is 2x+4/3 x^{3/2}+4+C