Integration by Partial Fraction Decomposition - Yahoo Answers

[MarkFL]

Here is the question:

I have problems solving this(integration by partial fractions)?


∫ (x^5 + 2x^3 - 3x)dx / (x^2 + 1)^3

I really don't know. Partial fractions. I know that you're supposed to do something like Ax+B when it comes to quadratic eqn's and you're supposed to give another arbitrary constant when it's raised to a power. but I don't think I can do it unless there's something else with it and I have no idea how to factor it. I really should've listened to our earlier algebra classes. :(

I have posted a link there so the OP can view my work.

 

[MarkFL]

Hello Vi3nc6en0t,

We are given to evaluate:

$$\int\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}\,dx$$

First, we observe that the numerator of the integrand can be factored:

$$x^5+2x^3-3x=x\left(x^4+2x^2-3 \right)=x\left(x^2+3 \right)\left(x^2-1 \right)=x(x+1)(x-1)\left(x^2+3 \right)$$

Thus, there are no common factors to divide out. To complete the partial fraction decomposition, we observe that an integrand with a repeated quadratic factor will decompose as follows:

$$\frac{P(x)}{\left(ax^2+bx+c \right)^n}=\sum_{k=1}^n\left(\frac{A_kx+B_k}{\left(ax^2+bx+c \right)^k} \right)$$

Thus, for the given integrand, we may write:

$$\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{\left(x^2+1 \right)^2}+\frac{Ex+F}{\left(x^2+1 \right)^3}$$

Multiplying through by $$\left(x^2+1 \right)^3$$, we obtain:

$$x^5+2x^3-3x=(Ax+B)\left(x^2+1 \right)^2+(Cx+D)\left(x^2+1 \right)+(Ex+F)$$

Expanding the right side and arranging on like powers of $x$, we obtain:

$$x^5+2x^3-3x=Ax^5+Bx^4+(2A+C)x^3+(2B+D)x^2+(A+C+E)x+(B+D+F)$$

Equating corresponding coefficients, we obtain the system:

$$A=1$$

$$B=0$$

$$2A+C=2\implies C=0$$

$$2B+D=0\implies D=0$$

$$A+C+E=-3\implies E=-4$$

$$B+D+F=0\implies F=0$$

Thus, we may state:

$$\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}=\frac{x}{x^2+1}-\frac{4x}{\left(x^2+1 \right)^3}$$

Now, in order to integrate, let's write:

$$\int\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}\,dx=\frac{1}{2}\int\frac{2x}{x^2+1}\,dx-2\int\frac{2x}{\left(x^2+1 \right)^3}\,dx$$

For both integrals, consider the substitution:

$$u=x^2+1\,\therefore\,du=2x\,dx$$

And we may now write:

$$\int\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}\,dx=\frac{1}{2}\int\frac{1}{u}\,du-2\int u^{-3}\,du$$

Applying the rules of integration, we obtain:

$$\int\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}\,dx=\frac{1}{2}\ln|u|-2\frac{u^{-2}}{-2}+C$$

Back substituting for $u$, and applying the property of logs that a coefficient may be taken inside as an exponent we have:

$$\int\frac{x^5+2x^3-3x}{\left(x^2+1 \right)^3}\,dx=\ln\left(\sqrt{x^2+1} \right)+\frac{1}{\left(x^2+1 \right)^2}+C$$