How do you integrate #1/ ((x + 1)(x + 2))# using partial fractions?

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Answer 1 · 2017-03-27T10:19:09

#int dx/((x+1)(x+2)) = ln abs ((x+1)/(x+2)) +C#

Write the rational function as a sum of partial fractions:

#1/((x+1)(x+2)) = A/(x+1)+B/(x+2)#

#1/((x+1)(x+2)) = (A(x+2)+B(x+1))/((x+1)(x+2))#

#1/((x+1)(x+2)) = (Ax+2A+Bx+B)/((x+1)(x+2))#

As the denominators are equal we can equate the numerators:

#1 = (A+B)x+2A+B#

and equating the coefficients with the same degree in #x# we have:

#{(A+B=0),(2A+B = 1):}#

#{(B=-A),(2A-A = 1):}#

#{(A=1),(B= -1):}#

So that:

#1/((x+1)(x+2)) = 1/(x+1)-1/(x+2)#

and integrating:

#int dx/((x+1)(x+2)) = int dx/(x+1)-int dx/(x+2)#

#int dx/((x+1)(x+2)) = ln abs (x+1) -ln abs (x+2) +C#

and using the properties of logarithms:

#int dx/((x+1)(x+2)) = ln abs ((x+1)/(x+2)) +C#