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

1. anonymous

2. anonymous

How the heck did wolfram jump from the first step to the second step? I feel like I am missing something.

3. anonymous

It's the second step I am not getting.

4. anonymous

$\int\limits\limits_{}^{}\frac{ x }{ (x^2-2x+5)^2 }dx$ I put that intergal into wolfram. Then I put the show steps.

5. anonymous

THe picture I posted is near the final answer.

6. anonymous

I see no other way to solve it.

7. anonymous

Eww... More partial fractions?

8. anonymous

But we only have 1 term divedby a polynomial . You have to use partial fractions.

9. anonymous

Eww...

10. anonymous

Thanks anyways.

11. anonymous

Partial fractions do not help at all. You can't reduce this.

12. anonymous

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.

13. anonymous

$\Large \sqrt{x^2-2x+5}=\sqrt{(x-1)^2 + 4}$ use trigonometric substitution $\Large x-1=2\tan \theta$

14. anonymous

@sirm3d : Okay why did you square root?

15. anonymous

oh, sorry, it was squared. the partial fraction is $\frac{x}{(x^2-2x+5)^2}=\frac{Ax+B}{x^2-2x+5}+\frac{Cx+D}{(x^2-2x+5)^2}$

16. anonymous

So it's actually possible to split this up? O_o .

17. TuringTest

yep

18. anonymous

I got A,B,D = 0. C=1.

19. anonymous

Is that right>

20. anonymous

?**

21. anonymous

@sirm3d

22. anonymous

i see that, too. $x=(1/2)(2x)=(1/2)(2x-2+2)=(1/2)(2x-2) + (1/2)(2)$

23. anonymous

$\frac{x}{(x^2-2x+5)^2}=\frac{(1/2)(2x-2)}{(x^2-2x+5)^2}+\frac{(1/2)(2)}{(x^2-2x+5)^2}$ 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 $x-1=2\tan\theta$

24. anonymous

So now I have: $\int\limits_{}^{}\frac{ 1 }{ (x^2-2x+5)^2 }dx$

25. anonymous

But we have a and b are 0 so the first term is entirely 0.

26. anonymous

i realized that too, partial fraction doesn't work. check my previous post.

27. anonymous

$x=Ax^3-2Ax^2+Bx^2-2Bx+Cx+5B+D$

28. anonymous

Yeah okay hmm...

29. anonymous

x-1=2tan(theta) dx=2sec^2(theta) d(theta)

30. anonymous

@sirm3d

31. anonymous

$\int\limits\limits_{}^{}\frac{ 1 }{ u^2 } du+ \int\limits\limits_{}^{}\frac{ 4\sec^2 \theta }{ 2(4\sec^2\theta) }d \theta$

32. anonymous

I get that which doesn't seem right...

33. anonymous

I used u instead of y.

34. anonymous

$x-1=2\tan\theta\\x^2-2x+5=4\sec^2\theta$ $\large (1/2)\int \frac{du}{u^2}+\int\frac{2\sec^2 \theta d\theta}{(2\sec \theta)^4}$

35. anonymous

$x-1=2\tan\theta$ $dx=2\sec ^2\theta d \theta$

36. anonymous

yes. our numerators agree.

37. anonymous

@amistre64 , can you take over? i have to catch some sleep.

38. anonymous

Thanks a lot :) .