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

Explanation

Here, the numbers `1^(2),2^(2),3^(2),4^(2),"...."i.e.1,4,9,16"...."` are not in AP but ` 1,4-1=3,9-4=5,16-9=7,"...."` are in AP.
Let `S_(infty)=1+2^(2)x+3^(2)x^(2)+4^(2)x^(3)+"...."" upto "infty`
`=1+4x+9x^(3)+16x^(3)+"...."" upto "infty "......(i)"`
Multiplying both sides of Eq. (i) by x, we get
`S_(infty)=x+4x^(2)+9x^(3)+16x^(4)+"...."" upto "infty "......(ii)"`
Subtracting Eq.(ii) from Eq.(i), we get
`(1-x)S_(infty)=1+3x+5x^(2)+7x^(3)+"...."" upto "infty "......(iii)"`
Again, multiplying both sides of Eq. (iii) by x, we get
`x(1-x)S_(infty)=x+3x^(2)+5x^(3)+7x^(4)+"...."" upto "infty "......(iv)"`
Subtracting Eq. (iv) from Eq. (iii),we get `(1-x)(1-x)S_(infty)=1+2x+2x^(2)+2x^(3)+"...."" upto "infty`
`=1+2(x+x^(2)+x^(3)+"...."" upto "infty)`
`=1+2((x)/(1-x))=((1+x)/(1-x))`
` therefore S_(infty)=((1+x))/((1-x)^(3))`