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I'm attempting to do the following integral

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

I wanted to try using partial fractions but I'm unsure how to factor the denominator. I've been unable to make any progress on this question because of this which is why I don't have any work for it. I was wondering whether there are any ways to take the integral of this function?

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  • $\begingroup$ Try writing numerator as sum of (product of some constant and derivative of denominator) and some constant $\endgroup$ – Atul Mishra Feb 13 '18 at 4:14
  • $\begingroup$ Letting $f(x) = x^3 + 2x^2 + 1,$ we find that $f(-2) = 1$ and $f(-3) = -8,$ so let us use $-2$ as a first approximation to the value of $a$ for which $f(a) = 0,$ and apply Newton's method. We get $a= -2.2055694.$ Thus we have $x^3 + 2x^2 + 1 = (x+2.205569)(x^2 + \cdots).$ Long division will give you the quadratic polynomial, and if its discriminant is negative, then the denominator has an irreducible quadratic factor. Then apply partial fractions. $\endgroup$ – Michael Hardy Feb 13 '18 at 5:15
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Hint

The denominator has three roots, only one neing real. So, write $$x^3+2x^2+1=(x-a)(x^2+bx+c)$$ and use partial fractions $$\frac{2x+1}{x^3 +2x^2 +1}=\frac 1 {a^+ab+c}\left(\frac{2 a+1}{x-a}-\frac{ (a+b-2 c)+(2 a+1) x}{x^2+bx+c} \right)$$

Now, work the problem of expressing $a,b,c$ which is not the most pleasant since $$a=-\frac{2}{3}-\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{43}{16}\right)\right)$$

Edit

If this looks too difficult, name $a,b,c$ the roots of $x^3 +2x^2 +1=0$ and write $$\frac{2x+1}{x^3 +2x^2 +1}=\frac{2x+1}{(x-a)(x-b)(x-c)}$$ Use partial fraction decomposition to get $$\frac{2x+1}{x^3 +2x^2 +1}=\frac{2 a+1}{(a-b) (a-c) (x-a)}+\frac{2 b+1}{(b-a) (b-c) (x-b)}+\frac{2 c+1}{(c-a) (c-b) (x-c)}$$ which is simple, leading to a weighted sum of logarithms, two of them with complex arguments (you can recombine them later). The result is ugly.

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    $\begingroup$ How did you go about expressing "a" in that form? $\endgroup$ – user262291 Feb 13 '18 at 4:31
  • $\begingroup$ @user262291. en.wikipedia.org/wiki/… $\endgroup$ – Claude Leibovici Feb 13 '18 at 4:32
  • $\begingroup$ Is it possible that there is no simpler solution? This is so complicated! I mean, when there is no obvious solution to the polynomial there should be a simpler solution rather then brute force. $\endgroup$ – Maffred Feb 13 '18 at 4:51
  • $\begingroup$ @Maffred. See my edit. $\endgroup$ – Claude Leibovici Feb 13 '18 at 5:02
  • $\begingroup$ I see, maybe there is no nice solution not involving these roots in an explicit way :( $\endgroup$ – Maffred Feb 13 '18 at 5:36

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