How do you use partial fraction decomposition to decompose the fraction to integrate $1/((x+2)(x^2+4))$ ?

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Answer 1 · 2015-10-24T01:11:49

$I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C$

$1/((x+2)(x^2+4)) = A/(x+2)+(Bx+C)/(x^2+4) =$

$= (A(x^2+4))/((x+2)(x^2+4))+((Bx+C)(x+2))/((x^2+4)(x+2)) =$

$= (Ax^2+4A+Bx^2+2Bx+Cx+2C)/((x+2)(x^2+4)) =$

$= ((A+B)x^2+(2B+C)x+4A+2C)/((x+2)(x^2+4))$

$(A+B)x^2+(2B+C)x+4A+2C = 1$

$A+B=0 => A=-B$
$2B+C=0 => C=-2B$
$4A+2C=1 => 4(-B)+2(-2B)=1$

$4(-B)+2(-2B)=1$

$-8B=1 => B=-1/8$

$A=-B => A=1/8$

$C=-2B => C=1/4$

$I = int 1/((x+2)(x^2+4)) dx = 1/8int 1/(x+2)dx + int (-1/8x+1/4)/(x^2+4)dx$

$I = 1/8ln|x+2| - 1/8 int (xdx)/(x^2+4) + 1/4int dx/(x^2+4)$

$I = 1/8ln|x+2| - 1/16 int ((x^2+4)'dx)/(x^2+4) + 1/4 *1/2arctan(x/2)$

$I = 1/8ln|x+2| - 1/16 int (d(x^2+4))/(x^2+4) + 1/8arctan(x/2)$

$I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +C$

How do you use partial fraction decomposition to decompose the fraction to integrate 1/((x+2)(x^2+4))1/((x+2)(x^2+4))?