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

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

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

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

- x - 1 - \frac{2}{x + 1} + C

Solution steps

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

Apply u-substitution

: \int \frac{- u ^{2} + 2}{u ^{2}} d u

= \int \frac{- u ^{2} + 2}{u ^{2}} d u

Expand

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

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

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 u + \int \frac{2}{u ^{2}} d u

\int 1 d u = u

\int \frac{2}{u ^{2}} d u = - \frac{2}{u}

= - u - \frac{2}{u}

Substitute back

u = x + 1

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

Simplify

= - x - 1 - \frac{2}{x + 1}

Add a constant to the solution

= - x - 1 - \frac{2}{x + 1} + C

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