How do you find $int (x-1)/((x+2)(x^2+3))dx$ using partial fractions?

Question

1619 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-02T09:19:43

The answer is $=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/(7sqrt3)arctan(x/sqrt3)+C$

Perform the decomposition into partial fractions

$(x-1)/((x+2)(x^2+3))=A/(x+2)+(Bx+C)/(x^2+3)$

$=(A(x^2+3)+(Bx+C)(x+2))/((x+2)(x^2+3))$

The denominators are the same, compare the numerators

$x-1=A(x^2+3)+(Bx+C)(x+2)$

Let $x=-2$ , $=>$ , $-3=7A$ , $=>$ , $A=-3/7$

Let $x=0$ , $=>$ , $-1=3A+2C$

$2C=-1-3A=-1+9/7=2/7$

$C=1/7$

Coefficients of $x^2$

$0=A+B$

$B=-A=3/7$

Therefore,

$(x-1)/((x+2)(x^2+3))=(-3/7)/(x+2)+((3/7)x+(1/7))/(x^2+3)$

So,

$int((x-1)dx)/((x+2)(x^2+3))=int(-3/7dx)/(x+2)+int(((3/7)x+(1/7))dx)/(x^2+3)$

$=-3/7ln(|x+2|)+3/14int(2xdx)/(x^2+3)+1/7int(1dx)/(x^2+3)$

$=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/21int(dx)/((x/sqrt3)^2+1)$

$=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/21*sqrt3arctan(x/sqrt3)+C$

$=-3/7ln(|x+2|)+3/14ln(x^2+3)+1/(7sqrt3)arctan(x/sqrt3)+C$

How do you find int (x-1)/((x+2)(x^2+3))dxint (x-1)/((x+2)(x^2+3))dx using partial fractions?