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+2x+2)^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 solved by integration by parts so,

`int uv'= uv-int u'v`

let` u= x-2 and v' = 1/((x^2+2x+2)^2)`

=>` u' = 1 `

so, `v=int 1/((x^2+2x+2)^2) dx`

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

let `u = x+1 , du =dx`

= `int 1/((u)^2+1)^2 du`

and now` u= tan(theta) so , du = (sec^2 (theta) ) d theta`

so ,

=>`int 1/((u)^2+1)^2 du`

=>`int 1/((tan(theta))^2+1)^2 (sec^2 (theta) ) d theta`

=>`int (sec^2 (theta) ) /((tan(theta))^2+1)^2 d theta`

=>`int (sec^2 (theta) ) /(sec^2(theta))^2 d theta`

=>`int cos^2 (theta) d theta`

=> we now that `int cos^2 t dt = (1/2)(t+(1/2)(sin(2t)))`

so ,

`int cos^2 (theta) d theta = (1/2)(theta+(1/2)(sin(2(theta))))`

Now ,

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

but `u= tan(theta) and x+1 = u`

so ,

V

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

now , as per `int uv' = uv -int u'v`

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

= `[(x-2)((1/2)(tan^(-1) (x+1) +(1/2)(sin(2(tan^(-1)(x+1))))))] `

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

now let us find the value of

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

=`int ((1/2)(tan^(-1) (x+1) )dx+int ((1/4)(sin(2(tan^(-1)(x+1)))))) dx`

= > let `tan^-1 (x+1) = u`

`tan u = x+1`

=> sec^2(u) = dx

so,

=`int ((1/2)(tan^(-1) (x+1) )+((1/4)(sin(2(tan^(-1)(x+1)))))) dx`

= =`int ((1/2)(u)+((1/4)(sin(2u)))) sec^2(u) du`

= `int (u/2 + sin(2u)/4 )sec^(2u) du `

=`1/2(utanu +ln(cosu))-1/4 ln(cos2u +1)`

= `1/2((x+1)(tan^(-1)(x+1)) +ln(cos(tan^(-1)(x+1))))-1/4 ln(cos2(tan^(-1)(x+1)) +1)`

so, now ,

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

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

`-1/2((x+1)(tan^(-1)(x+1)) +ln(cos(tan^(-1)(x+1))))-1/4 ln(cos2(tan^(-1)(x+1)) +1)`