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,

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

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