`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+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...
(The entire section contains 437 words.)
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