Completing the square and then using trig substitutio

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SUMMARY

The discussion centers on solving the integral of 1/[(x^2 + 4x + 3)^(3/2)] using the method of completing the square followed by trigonometric substitution. The correct approach involves rewriting the expression as 1/[(x + 2)^2 - 1]^(3/2) and substituting x + 2 = sec(theta), with dx = sec(theta)tan(theta). A critical error identified was neglecting to account for the exponent of 3/2 in the denominator, which significantly impacts the solution. The final result should not include a natural logarithm, indicating a need for careful attention to detail in the integration process.

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Homework Statement


Integral of 1/[(x^2+4x+3)^(3/2)]

The Attempt at a Solution


I tried completing the square and then using trig substitution. So its:

1/[(x+2)^2 -1] let x+2=sec(theta), dx =sec(theta)tan(theta)

After some simplifying, I get it down to integral of csc(theta) which = -ln|csc(theta) + cot(theta)|

My calc teacher has the answers online, and it doesn't have any ln in it at all, so I am pretty sure I am on the wrong track here. Any suggestions?
 
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It seems like you completely forgot about the exponent in the denominator.
 


KevinL said:

Homework Statement


Integral of 1/[(x^2+4x+3)^(3/2)]

The Attempt at a Solution


I tried completing the square and then using trig substitution. So its:

1/[(x+2)^2 -1] let x+2=sec(theta), dx =sec(theta)tan(theta)

After some simplifying, I get it down to integral of csc(theta) which = -ln|csc(theta) + cot(theta)

The small mistake you made after completing the square is that you forgit to copy down the ^3/2 in the denominator. It should be:

\frac{1}{[(x+2)^{2} -1]^{\frac{3}{2}}}.

It becomes much easier when you include it ;). But aside from that mistake you're on the right general track.
 
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