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1.19 Dealing with double roots

Example.

To express the following function as partial fractions, and integrate:

3⁢x-1(x+2)2{{3x-1}\over{(x+2)^{2}}}

Solution. We consider

3⁢x-1(x+2)2=Ax+2+B(x+2)2{{3x-1}\over{(x+2)^{2}}}={{A}\over{x+2}}+{{B}\over{(x+2)^{2}}}

for coefficients AA and BB to be determined; note that we need x+2x+2 as well as (x+2)2(x+2)^{2}. We cross multiply and find 3⁢x-1=A⁢(x+2)+B=A⁢x+(B+2⁢A),3x-1={A(x+2)+B}\,{=Ax+(B+2A),} hence A=3A={3} and B=-7.B={-7.} Now ∫3⁢x-1(x+2)2⁢d⁢x=∫(3x+2-7(x+2)2)⁢d⁢x=3⁢log⁡|x+2|+7x+2+c.\int{{3x-1}\over{(x+2)^{2}}}\,dx={\int\left({{3}\over{x+2}}-{7\over{(x+2)^{2}}% }\right)\,dx}\,{=3\log|x+2|+{7\over{x+2}}+c.}