Question

Someone want to explain this to me?

Answer 1

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

Answer 2

It's the second step I am not getting.

Answer 3

\[\int\limits\limits_{}^{}\frac{ x }{ (x^2-2x+5)^2 }dx\]

I put that intergal into wolfram.

Then I put the show steps.

Answer 4

THe picture I posted is near the final answer.

Answer 5

I see no other way to solve it.

Answer 6

Eww... More partial fractions?

Answer 7

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

Answer 8

Eww...

Answer 9

Thanks anyways.

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

@sirm3d : Okay why did you square root?

Answer 14

oh, sorry, it was squared.

the partial fraction is \[\frac{x}{(x^2-2x+5)^2}=\frac{Ax+B}{x^2-2x+5}+\frac{Cx+D}{(x^2-2x+5)^2}\]

Answer 15

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

Answer 16

yep

Answer 17

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

Answer 18

Is that right>

Answer 19

?**

Answer 20

@sirm3d

Answer 21

i see that, too.
\[x=(1/2)(2x)=(1/2)(2x-2+2)=(1/2)(2x-2) + (1/2)(2)\]

Answer 22

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

Answer 23

So now I have:

\[\int\limits_{}^{}\frac{ 1 }{ (x^2-2x+5)^2 }dx\]

Answer 24

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

Answer 25

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

Answer 26

\[x=Ax^3-2Ax^2+Bx^2-2Bx+Cx+5B+D\]

Answer 27

Yeah okay hmm...

Answer 28

x-1=2tan(theta)

dx=2sec^2(theta) d(theta)

Answer 29

@sirm3d

Answer 30

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

Answer 31

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

Answer 32

I used u instead of y.

Answer 33

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

Answer 34

\[x-1=2\tan\theta\]

\[dx=2\sec ^2\theta d \theta\]

Answer 35

yes. our numerators agree.

Answer 36

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

Answer 37

Thanks a lot :) .