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I have this integral:

\begin{equation*} \int \! \frac{x-1}{(x+1)(x^2+9)} \, \mathrm{d}x. \end{equation*}

I already split the denominator into two factors. Now, when I do partial fraction decomposition, I have this equation: $x-1=a(x^2+9)+b(x+1)$. I can easily set $x=-1$ and calculate $a=-\frac{1}{5}$. But how do I get $b$? I have tried putting the $a$ in and some random $x\neq -1$, but for each random $x$ I get a different value for $b$. Why is that so? And what is the proper method for calculating $b$?

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    $\begingroup$ You want $$x-1=a(x^2+9)+(bx+c)(x+1)$$ since you have that $x^2$ appearing in the denominator $\endgroup$
    – graydad
    May 5, 2015 at 17:59
  • $\begingroup$ The numerator has degree $<$ degree of the irreducible factor in denominator, hence for the case of $x^2+9$, it has degree possibly $1$. $\endgroup$
    – Bernard
    May 5, 2015 at 17:59

2 Answers 2

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You have to have $$x-1=a(x^2+9)+(bx+c)(x+1)$$ to get $$a=-\frac 15,\ b=\frac 15,\ c=\frac 45.$$ So, you'll have $$\frac{x-1}{(x+1)(x^2+9)}=\frac{-\frac{1}{5}}{x+1}+\frac{\frac 15x+\frac 45}{x^2+9}.$$

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The answer by mathlove hits the nail on the head. Alternatively, you could factor $x^2+9$ using complex numbers to get $(x-3i)(x+3i)$ which means you'd have $$x-1 = a(x-3i)(x+3i)+b(x+1)(x+3i)+c(x+1)(x-3i)$$ The numbers on the LHS are strictly real, so you'd know that the entire complex quantity on the RHS would be equal to zero. This isn't the easiest way to do it, but it will work and you can get some practice with complex numbers ;)

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