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I am trying to solve the following integral and would appreciate some guidance:

$$\int \frac{x + 1}{(x^2 + 6x + 14)^3} \, dx$$

I have attempted various methods, including substitution and integration by parts. One of the approaches I tried was substituting $x + 3$ into the expression, hoping to simplify the denominator. Despite these attempts, I am still unsure of how to proceed further. Could someone please provide a detailed explanation or point me in the right direction?

Thank you in advance for your help!

Gonçalo
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Math124
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  • Maple says $${\frac {-4,x-22}{40, \left( {x}^{2}+6,x+14 \right) ^{2}}}-{\frac {6 ,x+18}{200,{x}^{2}+1200,x+2800}}-{\frac {3,\sqrt {5}}{500}\arctan \left( {\frac { \left( 2,x+6 \right) \sqrt {5}}{10}} \right) }$$ – GEdgar Jul 19 '24 at 20:52

1 Answers1

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Hint: Complete the square in the denominator, manipulate the numerator to match it, then split the integral into two separate integrals. You will have two cases: one integral that is an straightforward substitution, and the other integral that you can use a trigonometric substitution ($x=a \tan \theta$). You can also use a hyperbolic substitution $x = a \sinh \theta$.

Note: If you choose $x = a \tan \theta$, since it's an indefinite integral, we are only concerned with positive values. Had the integral been a definite integral, we would require $x = a |\tan \theta|$ to determine what quadrant $\tan \theta$ lies. The hyperbolic subsitution $x = a \sinh \theta$ does not have that requirement.

Warning: the resulting integral will not be nice.

bjcolby15
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  • Thank you for the hint! However, I am concerned that $$ ( \tan(\theta) )$$ isn't monovalent, which might complicate the substitution. Could you please elaborate on how to handle this issue or suggest an alternative approach? – Math124 Jul 19 '24 at 19:06
  • @Math124 : Tangent is monovoalent; there are multiple choices of arctangent. However, bijectivity of the substition is not uniformly required. I discuss this at length here: https://math.stackexchange.com/a/3842386/123905 – Eric Towers Jul 19 '24 at 20:49
  • If by monovalent you mean "do we require a positive or negative value," we would require it if the integral were definite; then we would have to have $x = a |\tan \theta|$. Since we're dealing with an indefinite integral, we're not concerned with positive or negative, so we can keep $\tan \theta$ as positive (i.e. we can drop the absolute value bars). – bjcolby15 Jul 20 '24 at 11:34