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anonymous
 3 years ago
Someone want to explain this to me?
anonymous
 3 years ago
Someone want to explain this to me?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How the heck did wolfram jump from the first step to the second step? I feel like I am missing something.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's the second step I am not getting.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{}^{}\frac{ x }{ (x^22x+5)^2 }dx\] I put that intergal into wolfram. Then I put the show steps.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0THe picture I posted is near the final answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I see no other way to solve it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Eww... More partial fractions?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But we only have 1 term divedby a polynomial . You have to use partial fractions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Partial fractions do not help at all. You can't reduce this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Allright. 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \sqrt{x^22x+5}=\sqrt{(x1)^2 + 4}\] use trigonometric substitution \[\Large x1=2\tan \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@sirm3d : Okay why did you square root?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, sorry, it was squared. the partial fraction is \[\frac{x}{(x^22x+5)^2}=\frac{Ax+B}{x^22x+5}+\frac{Cx+D}{(x^22x+5)^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So it's actually possible to split this up? O_o .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got A,B,D = 0. C=1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i see that, too. \[x=(1/2)(2x)=(1/2)(2x2+2)=(1/2)(2x2) + (1/2)(2)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x}{(x^22x+5)^2}=\frac{(1/2)(2x2)}{(x^22x+5)^2}+\frac{(1/2)(2)}{(x^22x+5)^2}\] for the first term, use the substitution \[y=x^22x+5,\quad dy=(2x2)dx\] for the second term, use the trigonometric substitution \[x1=2\tan\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So now I have: \[\int\limits_{}^{}\frac{ 1 }{ (x^22x+5)^2 }dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But we have a and b are 0 so the first term is entirely 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i realized that too, partial fraction doesn't work. check my previous post.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x=Ax^32Ax^2+Bx^22Bx+Cx+5B+D\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x1=2tan(theta) dx=2sec^2(theta) d(theta)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{}^{}\frac{ 1 }{ u^2 } du+ \int\limits\limits_{}^{}\frac{ 4\sec^2 \theta }{ 2(4\sec^2\theta) }d \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I get that which doesn't seem right...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I used u instead of y.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x1=2\tan\theta\\x^22x+5=4\sec^2\theta\] \[\large (1/2)\int \frac{du}{u^2}+\int\frac{2\sec^2 \theta d\theta}{(2\sec \theta)^4}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x1=2\tan\theta\] \[dx=2\sec ^2\theta d \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. our numerators agree.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 , can you take over? i have to catch some sleep.
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