Inverse Trig Integration - Can I use u-substitution?

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SUMMARY

The discussion centers on evaluating an integral using u-substitution, specifically for the expression involving the numerator derived from the denominator. The key technique involves adding and subtracting 6 in the numerator to express it as the complete derivative of the denominator, \(x^2 + 6x + 13\). This allows for the substitution \(u = x^2 + 6x + 13\), simplifying the integral. The method is confirmed as the standard approach for this type of problem.

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Here is the problem and the solution manual's solution to the problem. I couldn't see that I had to add 6 and subtract 6 in the numerator then split the integral into two separate integrals. Is there an easier way to do it using u-substitution? Are their alternative ways to evaluate this integral?

http://img466.imageshack.us/img466/3753/problem0lw.jpg

Thanks
 
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From my experience, adding and subtracting 6 is the standard and easiest way to proceed.

The clue to add and subtract 6 is that, you want to express the numerator as the complete derivative of the denominator.

ie,
[tex]\frac{d}{dx} (x^2 + 6x +13) = 2x + 6[/tex]
Which enables you to use the substitution [itex]x^2 + 6x +13 = u[/itex].

The second part, ie, [tex]-6 \int \frac{1}{x^2 + 6x +13} dx[/tex] is then a standard integral.
 

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