How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$?

Question

Please could anyone help me to integrate $\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$ I know how to use partial fraction and I did this: $$ x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right) $$ And then ?.$\quad$ Thanks all.

Answer 1

We have:

$$ \int{\frac{4x+4}{x^4+x^3+2x^2}} dx $$

We need to decompose this fraction into pieces and then integrate each one separately. We start by simplifying the denominator:

$$ \frac{4x+4}{x^2(x^2+x+2)} $$

Since we have two quadratic terms in the denominator, we can guess the form: $$ \frac{4x+4}{x^2(x^2+x+2)} = \frac{Ax+B}{x^2} + \frac{Cx+D}{x^2+x+2} $$

Simplifying this further yields: $$ 4x+4 = (Ax+B)(x^2+x+2) + x^2(Cx+D) $$

Let's expand everything out so that we can solve for these constants A, B, C, and D:

$$ 4x+4 = (Ax^3+Ax^2+2Ax+Bx^2+Bx+2B)+(Cx^3+Dx^2) $$

$$ 4x+4 = (A+C)x^3+(A+B+D)x^2+(2A+B)x+2B $$

We now have a system of equations:

$$ A+C=0, A+B+D=0, 2A+B=4, 2B=4 $$

Solving (by working with the 2B=4 term first), we get: $$ A=1, B=2, C=-1, D=-3$$

Plugging back our constants into our "guess form": $$\frac{4x+4}{x^2(x^2+x+2)}=\frac{(1)x+(2)}{x^2}+\frac{(-1)x+(-3)}{x^2+x+2}$$ $$=\frac{x+2}{x^2} + \frac{-x-3}{x^2+x+2}$$ $$=\frac{1}{x}+\frac{2}{x^2}-\frac{x-3}{x^2+x+2} $$

That's much better, now you should be able to integrate each piece separately.

Can you finish?

Answer 2

Hint: $$ \begin{align} \frac{4x+4}{x^4+x^3+2x} &=\frac{4x+4}{x^2(x^2+x+2)}\\ &=\frac1x+\frac2{x^2}-\frac{x+1/2}{x^2+x+2}-\frac{5/2}{x^2+x+2} \end{align} $$

Answer 3

Your idea is good, write

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

Then, partial fraction will be of the form

$$\frac{x+1}{x^2 (x^2+x+2)}=\frac{a}{x} + \frac{b}{x^2} + \frac{cx+d}{x^2+x+2}$$

There are several methods to identify coefficients from this. The simplest IMHO is to rewrite the right hand side with the same denominator:

$$\frac{ax+b}{x^2} + \frac{cx+d}{x^2+x+2} = \frac{(ax+b)(x^2+x+2) + (cx+d)x^2}{x^2(x^2+x+2)}$$

Then develop, and write that the numerator must be $x+1$, thus you get equations by writing that the coefficients of $x^3$ and $x^2$ are $0$, and the coefficient of $x$ and the constant are $1$. Four linear equations, fours unknowns.

Is it enough for you to complete the process?