How do you integrate #4 /((x + 2)(x + 3))#?

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Answer 1 · 2018-03-20T06:52:57

The answer is #=4(ln(|x+2|)-ln(|x+3|))+C#

Perform the decomposition into partial fractions

#(4)/((x+2)(x+3))=A/(x+2)+B/(x+3)=(A(x+3)+B(x+2))/((x+2)(x+3))#

The denominators are the same, compare the numerators

#4=A(x+3)+B(x+2)#

Let #x=-2#, #=>#, #4=A#

Let #x=-3#, #=>#, #4=-B#

Therefore,

#(4)/((x+2)(x+3))=4/(x+2)+(-4)/(x+3)#

#int(4dx)/((x+2)(x+3))=int(4dx)/(x+2)+int(-4dx)/(x+3)#

#=4ln(|x+2|)-4ln(|x+3|)+C#