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I have the following integral :

$$\int \frac{x^2+2x+3}{x^2-5x+6}~dx$$

Although I tried to solve it by partial fractions, I could not come up with an appropriate answer. My question is more about the technique which I should use here in order to evalue the integral. I understand that the denominator could be expressed as $$(x-2)(x-3)$$ But then how to proceed with such a complicated numerator.

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  • $\begingroup$ Then $\int \frac{1}{x^2+x+2}dx$ = $\int \frac{1}{(x+1)^2+1}d(x+1)= tg^{-1} (x+1)$ $\endgroup$ – kmitov Mar 2 '16 at 7:09
  • $\begingroup$ Yeah, this is ok. But I have also an integral where this des not work. My question was more about the technique. I am editing the answer in order to explain better. Thenk you for the fast answers!! $\endgroup$ – S.19LaBG Mar 2 '16 at 7:29
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First, $$ x^2+2x+3=1\cdot (x^2-5x+6)+7x-3. $$ Thus $$ \frac{x^2+2x+3}{x^2-5x+6}=1+\frac{7x-3}{x^2-5x+6}. $$ You know $x^2-5x+6=(x-2)(x-3)$. It is known that there exists $A,B$ such that $$ \frac{7x-3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3} $$ and so $$ \frac{7x-3}{x^2-5x+6}=\frac{(A+B)x-(3A+2B)}{x^2-5x+6}. $$ Thus we get $A=-11$ and $B=18$. Therefore $$ \int\frac{x^2+2x+3}{x^2-5x+6}dx=x-11\ln|x-2|+18\ln|x-3|+C. $$

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  • $\begingroup$ Perfect answer. $\endgroup$ – S.19LaBG Mar 2 '16 at 7:55

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