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Evaluate: `int("x"^2+1)/(("x"+1)^2)"dx"`

Answer» `int (x^2+1)/(x+1)^2 dx = int (x^2+1+2x)/(x+1)^2 dx - int (2x)/(x+1)^2 dx`
`= int (x+1)^2/(x+1)^2 dx - int (2x)/(x+1)^2 dx`
`= int dx - int (2x)/(x+1)^2 dx`
`= int dx - int (2(x+1))/(x+1)^2 dx + int 2/(x+1)^2 dx`
Let `(x+1)^2 = t => 2(x+1)dx = dt`
Then, our integral becomes,
`= int dx - int dt/t + int 2/(x+1)^2 dx`
`= x -logt -2/(x+1)+c`
`= x-log(x+1)^2-2/(x+1)+c`
`= x-2log(x+1)-2/(x+1)+c`, which is the required value of given integral.


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