1/((x+2)(x^2+4)) = A/(x+2)+(Bx+C)/(x^2+4) =1(x+2)(x2+4)=Ax+2+Bx+Cx2+4=
= (A(x^2+4))/((x+2)(x^2+4))+((Bx+C)(x+2))/((x^2+4)(x+2)) ==A(x2+4)(x+2)(x2+4)+(Bx+C)(x+2)(x2+4)(x+2)=
= (Ax^2+4A+Bx^2+2Bx+Cx+2C)/((x+2)(x^2+4)) = =Ax2+4A+Bx2+2Bx+Cx+2C(x+2)(x2+4)=
= ((A+B)x^2+(2B+C)x+4A+2C)/((x+2)(x^2+4))=(A+B)x2+(2B+C)x+4A+2C(x+2)(x2+4)
(A+B)x^2+(2B+C)x+4A+2C = 1(A+B)x2+(2B+C)x+4A+2C=1
A+B=0 => A=-BA+B=0⇒A=−B
2B+C=0 => C=-2B2B+C=0⇒C=−2B
4A+2C=1 => 4(-B)+2(-2B)=14A+2C=1⇒4(−B)+2(−2B)=1
4(-B)+2(-2B)=14(−B)+2(−2B)=1
-8B=1 => B=-1/8−8B=1⇒B=−18
A=-B => A=1/8A=−B⇒A=18
C=-2B => C=1/4C=−2B⇒C=14
I = int 1/((x+2)(x^2+4)) dx = 1/8int 1/(x+2)dx + int (-1/8x+1/4)/(x^2+4)dxI=∫1(x+2)(x2+4)dx=18∫1x+2dx+∫−18x+14x2+4dx
I = 1/8ln|x+2| - 1/8 int (xdx)/(x^2+4) + 1/4int dx/(x^2+4)I=18ln|x+2|−18∫xdxx2+4+14∫dxx2+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