Re-express $\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}$ in terms of partial fractions and compute the indefinite integral $\dfrac{1}5{}\int f(x)dx $ using the result from the first part of the question.



Use $$\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}=\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F$$ and solve for $A,B,C,D,E$ and $F$.


Note that this partial fraction decomposition is a case where the degree of the numerator and denominator are the same. (Just notice that the sum of the highest degree terms in the denominator is $2+2+1=5.$) This means that the typical use of partial fraction decomposition does not apply.

Furthermore, notice that $(x^2+2x+1)$ factorizes as $(x+1)^2$, which means the denominator is really $(x+1)^3(2x^2-x+4)$. This means the most basic decomposition would involve denominators of $(x+1)^3$ and $(2x^2-x+4)$. However, we can escape that by using a more complicated decomposition involving the denominators $(x+1)^3$, $(x+1)^2$, $(x+1)$, and $(2x^2-x+4)$. The $F$ term is necessary for the $6x^5$ term to arise in the multiplication, intuitively speaking.

More thoroughly stated, the $F$ term is needed because of the following equivalency between $6x^5+x^2+x+2$ and the following expression: $$ \begin{align} &A(2x^2-x+4)\\ +&B(2x^2-x+4)(x+1)\\ +&C(2x^2-x+4)(x+1)^2\\ +&[Dx+E](x+1)^3\\ +&F(2x^2-x+4)(x+1)^3. \end{align} $$

Notice how the term $F(2x^2-x+4)(x+1)^3$ is the only possible term that can give rise to $6x^5$? That is precisely why it is there.

Hint 2

Part 1

In the integration of $\frac{1}{5}\int f(x)dx$, first separate the integral, $\int f(x)dx$ into many small integrals with the constants removed from the integrals. That is, $\int f(x) dx$ is $$A\int \frac{1}{(x+1)^3}dx+B\int \frac{1}{(x+1)^2}dx+C\int \frac{1}{x+1}dx+D\int \frac{x}{2x^2-x+4}dx\quad+E\int \frac{1}{2x^2-x+4}dx+F\int 1dx.$$

Part 2

Next, use the substitution $u=x+1$ with $du=dx$ on the first three small integrals:

$$A\int \frac{1}{u^3}du+B\int \frac{1}{u^2}du+C\int \frac{1}{u}du+D\int \frac{x}{2x^2-x+4}dx+E\int \frac{1}{2x^2-x+4}dx\quad +F\int 1dx.$$

Part 3

To deal with the integral $\int \frac{1}{2x^2-x+4}dx$, you must complete the square. This is done as follows:

$$ \begin{align} \int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\ &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx. \end{align} $$

To conclude, you make the trig substitution $x-\frac{1}{4}=\frac{\sqrt{31}}{4}\tan \theta$ with $dx=\left(\frac{\sqrt{31}}{4}\tan \theta+\frac{1}{4}\right)'d\theta=\frac{\sqrt{31}}{4}\sec^2\theta d\theta$. This gives you:

$$\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx=\frac{1}{2} \int \frac{\frac{\sqrt{31}}{4}\sec^2 \theta}{\frac{31}{16}\tan^2\theta+\frac{31}{16}}d\theta.$$

  • $\begingroup$ The degree of the numerator and denominator is the same; so you have to do the division first. Also note $x^2+2x+1=(x+1)^2$; so your factorization is incorrect. $\endgroup$ – David Mitra Dec 1 '12 at 20:10
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    $\begingroup$ There is a little problem with what you wrote. The remark of @DavidMitra regards the initial fraction - there the degree of the numerator and denominator is the same, which implies that there should be some constant factor, i.e: $$\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)}+\frac{Dx+E}{2x^2-x+4}+F$$ $\endgroup$ – Dennis Gulko Dec 1 '12 at 20:26
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    $\begingroup$ @Limitless Note that if you simplify the right hand side of the equation in your last comment, the degree of the numerator will be $4$ (and not $5$ as it should be). Generally, for the Partial Fraction decomposition method (as given here for example) to succeed, the degree of the numerator must be smaller than the degree of the denominator. $\endgroup$ – David Mitra Dec 1 '12 at 20:28
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    $\begingroup$ @Limitless: look at my edited remark. If you expand what you wrote, you get a polynomial of degree $4$ in the numerator. $\endgroup$ – Dennis Gulko Dec 1 '12 at 20:30
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    $\begingroup$ @Limitless The usual algorithm taught would give the wrong decomposition. However, you can correct it as Dennis writes in his comment above (in this case by adding a constant term $F$). Things will be ok then. But note the $F$ is added for this particular case. For other rational functions, a different correction term might be needed. It would be easier, in my opinion, just to do the division (polynomial division) in these cases where the degree of the numerator is at least the degree of the denominator. $\endgroup$ – David Mitra Dec 1 '12 at 20:35

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