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

The integral of (x^2+2x)/(sqrt(1-x)) is -2/5 (-x+1)^{5/2}+8/3 (-x+1)^{3/2}-6sqrt(-x+1)+C

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

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

- \frac{2}{5} (- x + 1)^{\frac{5}{2}} + \frac{8}{3} (- x + 1)^{\frac{3}{2}} - 6 - x + 1 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{1 - x} d x = - 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

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

= - 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) - 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

Simplify

- 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) - 4 (- \frac{1}{3} (1 - x)^{\frac{3}{2}} + 1 - x) : - \frac{2}{5} (- x + 1)^{\frac{5}{2}} + \frac{8}{3} (- x + 1)^{\frac{3}{2}} - 6 - x + 1

= - \frac{2}{5} (- x + 1)^{\frac{5}{2}} + \frac{8}{3} (- x + 1)^{\frac{3}{2}} - 6 - x + 1

Add a constant to the solution

= - \frac{2}{5} (- x + 1)^{\frac{5}{2}} + \frac{8}{3} (- x + 1)^{\frac{3}{2}} - 6 - x + 1 + C

What is the integral of (x^2+2x)/(sqrt(1-x)) ?
The integral of (x^2+2x)/(sqrt(1-x)) is -2/5 (-x+1)^{5/2}+8/3 (-x+1)^{3/2}-6sqrt(-x+1)+C