## Dido525 3 years ago Someone want to explain this to me?

1. Dido525

2. Dido525

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

3. Dido525

It's the second step I am not getting.

4. Dido525

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

5. Dido525

THe picture I posted is near the final answer.

6. Dido525

I see no other way to solve it.

7. Dido525

Eww... More partial fractions?

8. Dido525

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

9. Dido525

Eww...

10. Dido525

Thanks anyways.

11. Dido525

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

12. Dido525

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. sirm3d

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

14. Dido525

@sirm3d : Okay why did you square root?

15. sirm3d

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. Dido525

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

17. TuringTest

yep

18. Dido525

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

19. Dido525

Is that right>

20. Dido525

?**

21. Dido525

@sirm3d

22. sirm3d

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

23. sirm3d

$\frac{x}{(x^2-2x+5)^2}=\frac{(1/2)(2x-2)}{(x^2-2x+5)^2}+\frac{(1/2)(2)}{(x^2-2x+5)^2}$ 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. Dido525

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

25. Dido525

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

26. sirm3d

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

27. Dido525

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

28. Dido525

Yeah okay hmm...

29. Dido525

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

30. Dido525

@sirm3d

31. Dido525

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

32. Dido525

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

33. Dido525

I used u instead of y.

34. sirm3d

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

35. Dido525

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

36. sirm3d

yes. our numerators agree.

37. sirm3d

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

38. Dido525

Thanks a lot :) .