# How to find the integral of a quotient of rational functions?

How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$

My attempt:

We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$

It is easy to find that $A=1$, $B=2$, and $C=-1$.

Therefore

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

Therefore: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$

The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$

Where is my mistake?

• The first step in partial-fractions problems is to make sure the degree of the numerator is less than the degree of the denominator; if not, you perform long division to get a polynomial quotient, plus a remainder which satisfies the degree constraint. If you just try to mechanically do partial fractions as you did, you can get a "solution" for the coefficients, but it'll be wrong. – John Hughes May 14 '16 at 12:10

Your $A,B,C$ are wrong. You can't write the expression in that form because the numerator won't have the correct degree.

If you combine the terms, you get $$\frac{A(x+1)+Bx(x+1) +Cx^2}{x^3+x^2}$$

Note that the numerator is of degree at most $2$.

Instead, just perform polynomial division to get a quotient and a remainder term. You can then do what you did, working with the fractional remainder.

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

Your long division is wrong, because:

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

You need to combine polynomial division and partial fractions. \begin{align} \frac{x^4+1}{x^3+x^2} &=x-1+\frac{x^2+1}{x^2(x+1)}\tag{1}\\ &=x-1+\frac2{x+1}-\frac1x+\frac1{x^2}\tag{2} \end{align} Explanation:
$(1)$: polynomial division
$(2)$: partial fractions

To break a rational polymomial expression into parts, the degree of the numerator must be less than the degree of the denominator.

This is not the case with $\dfrac{x^4+1}{x^3+x^2}$. Using long division, we find

\begin{array}{rcccccccc} & & x & - & 1 &\\ & & --- & --- & --- & --- & --- & --- & ---\\ x^3 + x^2 & | & x^4 & + & 0x^3 & + & 0x^2 & + & 1 \\ & & --- & --- & --- \\ & & x^4 & + & x^3\\ & & --- & --- & --- & --- & --- \\ & & & & -x^3 & + & 0x^2 \\ & & & & -x^3 & - & x^2 \\ & & & & --- & --- & --- \\ & & & & & & x^2 & + & 1. \\ \end{array}

So $\dfrac{x^4+1}{x^3+x^2} = x - 1 + \dfrac{x^2 + 1}{x^3 + x^2}$.

And you need to solve $\dfrac{x^2+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$

$x^2 + 1 = A(x+1) + Bx(x+1) + Cx^2$

Let $x = -1$ and you get

$C = 2$

Let $C = 2$ and you get

$x^2 + 1 = A(x+1) + Bx(x+1) + 2x^2$

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

$1-x = A + Bx$

$A = 1$ and $B = -1$.

So $\dfrac{x^4+1}{x^3+x^2} = x - 1 + \dfrac{1}{x^2} - \dfrac{1}{x} + \dfrac{2}{x+1}$.

etc

Your A, B and C are wrong. Assuming your A,B and C are right, the integraiotn is right