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Mathematics
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How the heck did wolfram jump from the first step to the second step? I feel like I am missing something.
It's the second step I am not getting.

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Other answers:

\[\int\limits\limits_{}^{}\frac{ x }{ (x^2-2x+5)^2 }dx\] I put that intergal into wolfram. Then I put the show steps.
THe picture I posted is near the final answer.
I see no other way to solve it.
Eww... More partial fractions?
But we only have 1 term divedby a polynomial . You have to use partial fractions.
Eww...
Thanks anyways.
Partial fractions do not help at all. You can't reduce this.
Allright. It's almost 4:30 AM here. If someone could leave an explanation I would GREATLY appreciate it once I wake up in the morning.
\[\Large \sqrt{x^2-2x+5}=\sqrt{(x-1)^2 + 4}\] use trigonometric substitution \[\Large x-1=2\tan \theta\]
@sirm3d : Okay why did you square root?
oh, sorry, it was squared. the partial fraction is \[\frac{x}{(x^2-2x+5)^2}=\frac{Ax+B}{x^2-2x+5}+\frac{Cx+D}{(x^2-2x+5)^2}\]
So it's actually possible to split this up? O_o .
yep
I got A,B,D = 0. C=1.
Is that right>
?**
i see that, too. \[x=(1/2)(2x)=(1/2)(2x-2+2)=(1/2)(2x-2) + (1/2)(2)\]
\[\frac{x}{(x^2-2x+5)^2}=\frac{(1/2)(2x-2)}{(x^2-2x+5)^2}+\frac{(1/2)(2)}{(x^2-2x+5)^2}\] for the first term, use the substitution \[y=x^2-2x+5,\quad dy=(2x-2)dx\] for the second term, use the trigonometric substitution \[x-1=2\tan\theta\]
So now I have: \[\int\limits_{}^{}\frac{ 1 }{ (x^2-2x+5)^2 }dx\]
But we have a and b are 0 so the first term is entirely 0.
i realized that too, partial fraction doesn't work. check my previous post.
\[x=Ax^3-2Ax^2+Bx^2-2Bx+Cx+5B+D\]
Yeah okay hmm...
x-1=2tan(theta) dx=2sec^2(theta) d(theta)
\[\int\limits\limits_{}^{}\frac{ 1 }{ u^2 } du+ \int\limits\limits_{}^{}\frac{ 4\sec^2 \theta }{ 2(4\sec^2\theta) }d \theta\]
I get that which doesn't seem right...
I used u instead of y.
\[x-1=2\tan\theta\\x^2-2x+5=4\sec^2\theta\] \[\large (1/2)\int \frac{du}{u^2}+\int\frac{2\sec^2 \theta d\theta}{(2\sec \theta)^4}\]
\[x-1=2\tan\theta\] \[dx=2\sec ^2\theta d \theta\]
yes. our numerators agree.
@amistre64 , can you take over? i have to catch some sleep.
Thanks a lot :) .

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