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

The integral of ((x-2)sqrt(1-x^2))/(1-x^2) is -sqrt(1-x^2)-2arcsin(x)+C

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

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

- 1 - x^{2} - 2 arcsin (x) + C

Solution steps

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

Simplify

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

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

Expand

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

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

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

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

= - 1 - x^{2} - 2 arcsin (x)

Add a constant to the solution

= - 1 - x^{2} - 2 arcsin (x) + C

\frac{d}{d x} ((3 x - 4) (5 x + 2))

\frac{\partial}{\partial x} (- \frac{y}{x ^{2} + y ^{2}})

y = \frac{1}{1 + x ^{2}}, (1, 0.5)

y = x^{3} + 6 x - 2, (1, 5)

\frac{d}{d x} (\frac{9}{6 - x})

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