What is the integral of 1/(x^2 + 4x + 3)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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