What is the integral of 1/(sqrt(x-1)-\sqrt{x)} ?

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

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integral of 1/(sqrt(x-1)-\sqrt{x)}

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

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

Solution steps

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

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

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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

x \to 5 + lim (\frac{- 3}{x ^{2} - 25})

(1 + x^{2}) y^{^{'}} = x (1 - y)

1 + 4 x - x^{2}, 11 - x - x^{2}, 1 - \frac{2 x}{3}

x + 1, 2 x - 4, 0, 3

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