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Jun 16 10 7:13 AM

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Please help me integrate this expression w.r.t. x ----

(x^2 + 1) ^-2

where ^ represents raise to the power.

Thanks in advance..
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#1 [url]

Jun 23 10 11:52 PM

Let x = tan(u)

dx = sec^2(u) du


It is then the integral ( 1/(1+tan^2(u))^2 * dx)

Substituting you get integral (1/(sec^2(u))^2 * sec^2(u) du)

Then you get int(1/(sec^2(u))) = int(cos^2(u))

Which you can look up or solve by parts. This gives you
(cos(u)sin(u) + u)/2

To find in terms of xs (use a triangle with sides x , 1, Sqrt(1+x^2))

You get (x/(1+x^2)) + tan^-1(x))/2

If you need further explanation post again.

image

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#2 [url]

Nov 28 11 9:09 AM

outqast wrote:
Let x = tan(u)

dx = sec^2(u) du


It is then the integral ( 1/(1+tan^2(u))^2 * dx)

Substituting you get integral (1/(sec^2(u))^2 * sec^2(u) du)

Then you get int(1/(sec^2(u))) = int(cos^2(u))

Which you can look up or solve by parts. This gives you
(cos(u)sin(u) + u)/2

To find in terms of xs (use a triangle with sides x , 1, Sqrt(1+x^2))

You get (x/(1+x^2)) + tan^-1(x))/2

If you need further explanation post again.


wow! why don't just educate students? it's as if you all know the math!

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