What is #f(x) = int (x^2+x)/((x+3)(x+4) ) dx# if #f(-2)=2 #?

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Answer 1 · 2017-12-12T11:30:52

#f(x)=x+4+12Ln(2)+6Ln(x+3)-12Ln(x+4)#

#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#

=#int (x^2+7x+12-6x-12)/[(x+3)*(x+4)]*dx#

=#int dx-int (6x+12)/[(x+3)*(x+4)]*dx#

=#x-int (6x+12)/[(x+3)*(x+4)]*dx+C#

Now I solved #int (6x+12)/[(x+3)*(x+4)]*dx# integral,

#(6x+12)/[(x+3)*(x+4)]=A/(x+3)+B/(x+4)#

After expanding denominator,

#A*(x+4)+B*(x+3)=6x+12#

Set #x=-4#, #-B=12#, so #B=12#

Set #x=-3#, #A=-6#

Hence,

#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#

=#x-[int -6/(x+3)*dx+12/(x+4)*dx]+C#

=#x-(-6Ln(x+3)+12Ln(x+4))+C#

=#x+6Ln(x+3)-12Ln(x+4)+C#

After using #f(-2)=2# condition,

#-2+6Ln(-2+3)-12Ln(-2+4)+C=2#

#C+6Ln(1)-12Ln(2)-2=2#

#C+0-12Ln(2)-2=2#

#C=4+12Ln2#

Thus,

#f(x)=x+4+12Ln(2)+6Ln(x+3)-12Ln(x+4)#