How do you integrate $(x+4)/((x+1)(x-2)^2 )$ using partial fractions?

Question

1381 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-22T12:17:13

The answer is $=1/3ln(|x+1|)-1/3ln(|x-2|)-2/(x-2)+C$

Perform the decomposition into partial fractions

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

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

The denominators are the same, compare the numerators

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

Let $x=-1$ , $=>$ , $3=9A$ , $=>$ , $A=1/3$

Let $x=2$ , $=>$ , $6=3C$ , $=>$ , $C=2$

Coefficients of $x^2$

$0=A+B$ , $=>$ , $B=-1/3$

Therefore,

$(x+4)/((x+1)(x-2)^2)=(1/3)/(x+1)+(-1/3)/(x-2)+(2)/(x-2)^2$

$int((x+4)dx)/((x+1)(x-2)^2)=int(1/3dx)/(x+1)+int(-1/3dx)/(x-2)+int(2dx)/(x-2)^2$

$=1/3ln(|x+1|)-1/3ln(|x-2|)-2/(x-2)+C$

How do you integrate (x+4)/((x+1)(x-2)^2 )(x+4)/((x+1)(x-2)^2 ) using partial fractions?