How do you use partial fraction decomposition to decompose the fraction to integrate 1/((x+2)(x^2+4))1(x+2)(x2+4)?

1 Answer
Oct 24, 2015

I = 1/8ln|x+2| - 1/16 ln(x^2+4) + 1/8arctan(x/2) +CI=18ln|x+2|116ln(x2+4)+18arctan(x2)+C

Explanation:

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=0A=B
2B+C=0 => C=-2B2B+C=0C=2B
4A+2C=1 => 4(-B)+2(-2B)=14A+2C=14(B)+2(2B)=1

4(-B)+2(-2B)=14(B)+2(2B)=1

-8B=1 => B=-1/88B=1B=18

A=-B => A=1/8A=BA=18

C=-2B => C=1/4C=2BC=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=181x+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|18xdxx2+4+14dxx2+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