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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\int x sec^{2} (5 x) d x

y^{^{''}} + 4 y = - 2

\int cos (φ) d φ

y^{'} = 0.027 y (1 - \frac{y}{1527})

ln (3) + e^{x^{e}}

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