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Find $$\int^u_0\frac{dx}{x^2+3x+2}$$

I'm having trouble making progress on this. Specifically, I know the following means to find an integral:

  1. Antiderivative
  2. Integration by parts
  3. Change of variable
  4. Definition of integral (either Darboux or Riemann).

None of these have been helpful to me:

  1. $\frac{1}{x^2 + 3x + 2}$ is not the derivative of anything I can construct. The closest is $\frac{2x + 3}{x^2 + 3x + 2} = [\log (x^2 + 3x + 2)]'$. This might be useful in combination with another method, so let's keep on looking.
  2. I've tried various combinations, such as $u = \frac{1}{x^2 + 3x + 2}, v= x^2 + 3x$ or $u = \log(x^2 + 3x + 2), v = \frac{1}{2x+3}$. Again, none have gotten me further.
  3. $f(x) = 1/x, u(x) = x^2 + 3x +2, u'(x) = 2x$. The problem is I can't get the $u'(x)$ factor to appear.
  4. I've made no progress with the definitions. In fact, I've only seen definitions used to prove general theorems, never to compute a specific integral.

Can you point me in the right direction? Please do not post the full solution; just let me know how I need to tackle this.

Blue
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SRobertJames
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    This is less of an integration technique and more of an algebraic manipulation, but try partial fraction decomposition: can you rewrite your integrand in terms of a sum of simpler fractions? (try factoring the denominator) – Stephen Donovan Jun 15 '21 at 22:16
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    Have you learned partial-fraction decomposition yet? The denominator can be factored and the single rational function can be separated into the sum of two simpler ones. –  Jun 15 '21 at 22:16
  • I think that if you complete the square the results follow – Brien Navarro Jun 15 '21 at 22:18
  • Read, at your liberty, this answer which covers partial fraction decomposition. You can use this to solve questions of the form $\int \frac{P(x)}{Q(x)}$. – Sarvesh Ravichandran Iyer Jun 15 '21 at 22:20
  • @BrienNavarro This should work but it's a lot messier than the partial fraction decomposition. I believe for this type of integral as a general rule of thumb completing the square and going for a trig sub is better when the roots in the denominator are complex. – Stephen Donovan Jun 15 '21 at 22:24
  • For fun you can let $u=\frac{x+1}{x+2}$ – kingW3 Jun 15 '21 at 22:42
  • I have closed your question as a duplicate. The dupe target is not a precise duplicate, but the relevant technique here is to decompose the integrand via partial fractions, then integrate. The linked duplicate demonstrates how this can be done. – Xander Henderson Jun 15 '21 at 22:44

2 Answers2

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HINT

You can factor the denominator in order to obtain:

\begin{align*} \frac{1}{x^{2} + 3x + 2} & = \frac{1}{(x + 1)(x + 2)}\\\\ & = \frac{(x+2) - (x+1)}{(x+1)(x+2)}\\\\ & = \frac{1}{x+1} - \frac{1}{x+2} \end{align*}

Can you take it from here?

user0102
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I will just provide the Ansatz for partial fraction decomposition.

You factor by looking for the roots of $x^2+3x+2$, which might be $x_1$ and $x_2$. I use completion of the square.

Then, if we got two different roots $x_i$, you set $$ \frac{1}{x^2+3x+2} = \frac{1}{(x-x_1)(x-x_2)} = \frac{C}{x-x_1} + \frac{D}{x-x_2} $$ and multiply both sides by $(x-x_1)(x-x_2)$ and sort by powers of $x^0 = 1$ and $x^1 = x$ to get a linear system of two equations in the unknowns $C$ and $D$.

After you determined $C$ and $D$ you can continue to integrate the easier two integrals.

mvw
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