6
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – John Hughes May 14 '16 at 12:10
5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

Your long division is wrong, because:

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.