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

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Answer 1 · 2015-12-03T02:19:35

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

Original
$int x^2/(x^2+3x+2)dx " " " " " " (1)$

Step 1: Rewrite the expression as a proper fraction (using long division)

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

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

Step 2: Factor the denominator $(3)$

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

Set up the partial fraction decomposition
$(3x+2)/(x^2 + 3x +2) =$

$=(3x+2)/((x+1)(x+2)) = color(red)A/(x+2) + color(red)B/(x+1) " " " " " " (4)$

Multiply LCD (least common denominator)

$(3x+2) = A(x+1) + B(x+2)$

Let $x= -1$
$(3*-1+2)= A(-1+1)+B(-1+2)$
$color(red)(-1= B)$

Let $x= -2$
$3*-2+2= A(-2+1)+B(-2+2)$
$-4 = -A$
$color(red)(4=A)$

Rewrite $(3)$
$color(red)4/(x+2) - color(red) (1)/(x+1)$

Rewrite (2)

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

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

$color(blue)( =x -4 ln|x+2| -ln|x+1| +C) " " " " " (5)$

I hope this help.

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