`int (x^3 + 2x^2 + 3x - 2)/(x^2 + 2x + 2)^2 dx` Evaluate the integral

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Given to solve

`int (x^3+2x^2+3x-2)/((x^2+2x+2)^2) dx`

but first we have to find this

`(x^3+2x^2+3x-2)/((x^2+2x+2)^2) `

= `(x^3+2x^2+2x+x-2)/((x^2+2x+2)^2) `

= `(x(x^2+2x+2)+x-2)/((x^2+2x+2)^2)`


= `x/(x^2+2x+2) +(x-2)/((x^2+2x+2)^2)`

`(x^3+2x^2+3x-2)/((x^2+2x+2)^2) = x/(x^2+2x+2) + (x-2)/((x^2+2x+2)^2) `

Integrating both sides we get:

`int (x^3+2x^2+3x-2)/((x^2+2x+2)^2) dx = int x/(x^2+2x+2) dx + int (x-2)/((x^2+2x+2)^2) dx`

first let us solve,

` int x/(x^2+2x+2) dx`


= `int (x+1-1)/((x+1)^2+1) dx`

`int (x+1)/((x+1)^2+1) dx +int (-1)/((x+1)^2+1) dx`

as we know

` int u/(u^ +1) dx= (ln(u^2+1))/2`


and `int 1/(u^2+1)dx=tan^(-1) u `

so ,

`int x/(x^2+2x+2) dx`

=`int (x+1)/((x+1)^2+1) dx -int (1)/((x+1)^2+1) dx`

=`(ln((x+1)^2+1))/2 -tan^(-1) (x+1) `



 and now we have to solve ,

` int (x-2)/((x^2+2x+2)^2) dx`


this can be...

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