The integral `int 1/(x^2 + 4x + 3) dx` has to be determined. Use partial fractions.
x^2 + 4x + 3
=> x^2 + 3x + x + 3
=> x(x + 3) + 1(x + 3)
=> (x + 1)(x + 3)
`1/(x^2 + 4x + 3)` = `A/(x + 1) + B/(x + 3)`
=> `1/(x^2+4x+3) = (Ax + 3A + Bx + B)/(x^2+4x+3)`
=> `Ax + 3A + Bx + B = 1`
This gives
A + B = 0...(1)
3A + B = 1 ...(2)
(2) - (1)
=> 2A = 1
=> A = 1/2
B = -1/2
The given integral can be written as
`int 0.5/(x + 1) - 0.5/(x + 3) dx`
=> `0.5*ln(x + 1) - 0.5*ln(x + 3) + C`
The integral `int 1/(x^2 + 4x + 3) dx` = `0.5*ln(x + 1) - 0.5*ln(x + 3) + C`
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