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From: Anolethron on 14 Mar 2005 13:04
What substitution do you think is convenient ? I tried arctanx=y but it
doesn't lead anywhere good.

Int [xe^(arctanx)]/{Sqrt[(1+x^2)^3]} dx ??


From: Stephen J. Herschkorn on 14 Mar 2005 13:59
Anolethron wrote:

>What substitution do you think is convenient ? I tried arctanx=y but it
>doesn't lead anywhere good.
>
>Int [xe^(arctanx)]/{Sqrt[(1+x^2)^3]} dx ??
>
>

Just an idea: Write the integrand as x/(1+x^2) * exp(arctan x) /
sqrt(1+x^2). The second factor is e^v dv/dx. Integrate by parts.

If this is homework, cite sources of assistance in submitted work.

--
Stephen J. Herschkorn sjherschko(a)netscape.net
From: Anolethron on 14 Mar 2005 14:54
This makes it even more difficult actually


From: Stephen J. Herschkorn on 14 Mar 2005 15:03
Stephen J. Herschkorn wrote:

> Anolethron wrote:
>
>> What substitution do you think is convenient ? I tried arctanx=y but it
>> doesn't lead anywhere good.
>>
>> Int [xe^(arctanx)]/{Sqrt[(1+x^2)^3]} dx ??
>>
>>
>
> Just an idea: Write the integrand as x/(1+x^2) * exp(arctan x) /
> sqrt(1+x^2). The second factor is e^v dv/dx. Integrate by parts.


I meant x / sqrt(1+x^2) * exp(arctan x) / (1 + x^2).


Anolethron wrote:

>This makes it even more difficult actually
>

I see what you mean. The integrand is actually du/dx * dv/dx,
where u = sqrt(1+x^2) and v = exp(arctan x). Offhand, I don't see
what to do with it. Quickmath (www.quickmath.com) gives you the
following answer.

-(e^arctan(x)*(1 + x^2))/(2*sqrt((1 + x^2)^3)) + (e^arctan(x)*x*(1 +
x^2))/(2*sqrt((1 + x^2)^3))


--
Stephen J. Herschkorn sjherschko(a)netscape.net
From: W. Dale Hall on 14 Mar 2005 15:38


Anolethron wrote:
> What substitution do you think is convenient ? I tried arctanx=y but it
> doesn't lead anywhere good.
>
> Int [xe^(arctanx)]/{Sqrt[(1+x^2)^3]} dx ??
>
>

I must be doing things wrong, then. If I substitute y = arctan(x), I
have dy = dx / (1+x^2), and the substitution looks like this:

Int [xe^(arctanx)]/{Sqrt[(1+x^2)^3]} dx

= Int [tan(y) e^y]/{Sqrt[(1+tan(y)^2)]} dy

which ( identifying sqrt(sec^2(y)) with sec(y) ) is this:

= Int [tan(y)/sec(y)] e^y dy

= Int [sin(y) e^y] dy.

The latter can be done any number of ways. Of course, you then
need to put everything back together, but that's not a big
problem.

The identification of sqrt(sec^2(y)) with sec(y) is of course
not entirely correct; properly, one should write |sec(y)| instead.
However, note that y = arctan(x), so y takes values in [-pi/2 pi/2],
where cos(y) > 0. In essence, since sec(y) never takes negative
values in the domain visited by y for the substitution, we can
replace |sec(y)| by sec(y).

Dale.
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