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

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

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

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

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

Solution steps

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

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

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

Simplify

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

= \int 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 1 d x + \int x d x

\int 1 d x = x

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial y} (x e^{w})

x \to 2 - lim (\frac{3 x}{x - 2})

f (x) = (x^{2} + 3 x)^{2}

\int \frac{8}{x ^{2} x ^{2} + 16} d x

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