# polynomial long division Integration

$$\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx$$ I've been trying to solve this integral, my gut tells me to use polynomial long division so I factored and got $$\int \frac{3x^2 +6x +2}{(2+x)(1+x)}dx$$ I have never done a division with two factors in the denominator, would anybody be able to help out?

• You can do long division without factoring the denominator Mar 8, 2018 at 23:15

$\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx$

Since $\dfrac1{x^2 + 3x +2} =\dfrac1{(x+2)(x+1)} =\dfrac1{x+1}-\dfrac1{x+2}$ and

$\begin{array}\\ \dfrac{3x^2 +6x +2}{x^2 + 3x +2} &=\dfrac{3(x^2+3x+2)-3(3x+2) +6x +2}{x^2 + 3x +2}\\ &=3-\dfrac{3x+4}{x^2 + 3x +2}\\ &=3-(3x+4)(\dfrac1{x+1}-\dfrac1{x+2})\\ &=3-\dfrac{3x+4}{x+1}-\dfrac{3x+4}{x+2}\\ &=3-\dfrac{3(x+1)+1}{x+1}+\dfrac{3(x+2)-2}{x+2}\\ &=3-3-\dfrac{1}{x+1}+3-\dfrac{2}{x+2}\\ &=3-\dfrac{1}{x+1}+\dfrac{2}{x+2}\\ \end{array}$

and this you can integrate by yourself.

Guide:

$$3x^2+6x+2=3(x^2+3x+2)-3x-4$$

$$\frac{3x^2+6x+2}{x^2+3x+2}=3- \frac{3x+4}{(2+x)(1+x)}$$

You might like to perform partial fraction on the second term to solve the problem.

Alt. hint: $\;3x^2 +6x +2=3(x^2+3x+2)-3x-4=3(x+1)(x+2)-2(x+1)-(x+2)$

You do polynomial long division just like you do long division on integers.

$\require{enclose} \begin{array}{r} 3 \\ x^2 + 3x + 2 \enclose{longdiv}{3x^2+6x+2} \\ \underline{3x^2 + 9x + 6} \\ -3x - 4 \end{array}$

$\frac {3x+6x + 2}{x^2+3x + 2} = 3 - \frac {3x+ 4}{x^2+3x+2}$

But I think what you are really looking for is something like

$\frac {3x^2+6x + 2}{x^2+3x + 2} = A + \frac {B}{x+1} + \frac {C}{x+2}\\ 3x^2+6x + 2 = A(x^2 + 3x + 2) + B(x+2) + C(x+1)\\ 3x^2 + 6x +2 = Ax^2 +(3A + B + C)x + (2A + 2B + C)$

The two polynomial are equal if all of the coefficients are equal

$A = 3\\ 3A + B + C = 6\\ 2A + 2B + C = 2$

And solve for $A,B,C$