Integration - problems with defining the original function

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kapitan90
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Homework Statement


Hi,
I am supposed to integrate [tex] \frac{1}{(x^2 + 2x +5)}[/tex]
In my textbook the method of arctan is used.

Homework Equations


I wonder why I can't define y as the logarithmic function instead.

The Attempt at a Solution



y = ln (x^2 + 2x +5) / (2x+2)

with absolute value in ln.Can both methods be used?
When I compare graphs for ln and arctan they are equivalent despite x close to the y - axis.
 
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kapitan90 said:

Homework Statement


Hi,
I am supposed to integrate [tex] \frac{1}{(x^2 + 2x +5)}[/tex]
In my textbook the method of arctan is used.

Homework Equations


I wonder why I can't define y as the logarithmic function instead.


The Attempt at a Solution


[tex] \frac{dy}{dx}= \frac{1}{x^2 + 2x +5()}[/tex]

y = ln (x^2 + 2x +5) / (2x+2)

with absolute value in ln.


Can both methods be used?

No. To take the derivative of y = ln (x^2 + 2x +5) / (2x+2) you need to use the quotient rule. The result isn't 1/(x^2+2x+5).
 
Ok, I see that it's incorrect. So can we generalize that to use direct reverse method with natural logarihm the integrated function needs to have a form:

[tex] <br /> \frac{af'(x)}{f(x)}<br /> [/tex]

and ln(f(x)) - method is basically invalid for all other forms?
 
Yes, that is the same as the standard integration formula:
[tex]\int \frac{du}{u}= ln(c)+ C[/tex]
with f(x) instead of u.

To integrate
[tex]\int \frac{dx}{x^2+ 2x+ 5}[/tex]
I suggest you complete the square in the denominator to get a standard form.
 
Ok, now I get it, thanks for help!