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I was wondering if someone would be kind enough to walk me through the logic used in solving the following integral. I have been to class, and have read the section (9.4 of Swokowski's Classic), and have studied the answer in the solution manual, but I can't quite seem to make sense of the rules posed (p.474, Swokowski's Classic) for the decomposition.

$$ \int\frac{x^2+3x+1}{x^4+5x^2+4}dx $$

Factors to:

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

And this is where I get completely lost. I can do simple ones, such as

$$ \int\frac{x+16}{x^2+2x-8} $$

where they reduce to

$$ \frac{A}{x+4} + \frac{B}{x-2} $$

But the solution manual suggests that A and B should be Ax+B and Cx+D, referring to the aforementioned rule, and I'm quite confused.

Thank you very much!

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Procedure of decomposition

So:

$$\frac{x^2+3x+1}{(x^2+4)(x^2+1)} = \frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+1} \Rightarrow$$

$$ \Rightarrow x^2+3x+1 =(x^2+1)(ax+b)+(x^2+4)(cx+d) \Rightarrow$$

$$\Rightarrow x^2+3x+1 =(a+c)x^3+(b+d)x^2+(a+4c)x+(b+4d)$$

Therefore , you have to solve following system of equations :

$\begin{cases} a+c=0 \\ b+d=1 \\ a+4c=3 \\ b+4d=1 \end{cases}$

After you find coefficients : $a,b,c,d$ use sum of integrals :

$$\int\frac{ax+b}{x^2+4} \,dx =\int\frac{ax}{x^2+4} \,dx+\int\frac{b}{x^2+4} \,dx$$

$$\int\frac{cx+d}{x^2+1} \,dx =\int\frac{cx}{x^2+1} \,dx+\int\frac{d}{x^2+1} \,dx$$

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You can still do the partial fraction decomposition for the integrand, but now $A$ and $B$ would need to be linear functions of $x$, rather than constants: $$ \frac{x^2+3x+1}{(x^2+4)(x^2+1)} = \frac{1-x}{x^2+4}+\frac{x}{x^2+1} $$ Then use table integrals: $$ \int \frac{a x+b}{x^2+d^2} \mathrm{d} x = \frac{b}{d} \arctan\left(\frac{x}{d} \right) + \frac{a}{2} \log\left(x^2+d^2\right) $$

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  • $\begingroup$ I'm still a bit confused about getting to the knowledge that A & B need to be linear functions, and, how you got to 1-x & x. $\endgroup$ – Josh Jan 29 '12 at 17:22

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