Integration - problems with defining the original function

[kapitan90]

Homework Statement

Hi,
I am supposed to integrate <br /> \frac{1}{(x^2 + 2x +5)}<br />
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.

 

[Dick]

Homework Statement

Hi,
I am supposed to integrate <br /> \frac{1}{(x^2 + 2x +5)}<br />
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

<br /> \frac{dy}{dx}= \frac{1}{x^2 + 2x +5()}<br />

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).

 

[kapitan90]

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:

<br /> <br /> \frac{af&#039;(x)}{f(x)}<br /> <br />

and ln(f(x)) - method is basically invalid for all other forms?

 

[HallsofIvy]

Yes, that is the same as the standard integration formula:
\int \frac{du}{u}= ln(c)+ C
with f(x) instead of u.

To integrate
\int \frac{dx}{x^2+ 2x+ 5}
I suggest you complete the square in the denominator to get a standard form.

 

[kapitan90]

Ok, now I get it, thanks for help!