Integration again (1 Viewer)

[Arowana21]

(2x-3)/(1-2x) dx

 

[Arowana21]

actually i know how to do it now..... here is another one

1/(x^2 +x +1)

 

[pLuvia]

Use complete the square then integrate and should get something in tan^-1(f(x))

 

[vds700]

actually i know how to do it now..... here is another one

1/(x^2 +x +1)

=I 1/[(x + 1/2)^2 + 3/4]
= 2/sqrt3 tan^-1(2x + 1)/sqrt3 + c

 

[Arowana21]

I have another one....

3/ (9x^2 - 6x +2)

 

[conics2008]

ok 3 S 1/ (x-1/3)+1/9

use u = x-1/3 du/dx = 1 >> du=dx

3S 1/u^2 +1/9

then when you do this

you get tan^-1 ( 3(u) )

sub u back in .. therefore you get

tan^-1(3x-1) + C

 

[Arowana21]

how about...

x/(x^2 + 2x -2)

 

[pLuvia]

how about...

x/(x^2 + 2x -2)

Try manipulating the numerator so that it is equal to 2x+2, then see if you can see the answer

 

[vds700]

how about...

x/(x^2 + 2x -2)

I x/(x^2 + 2x - 2)
= (1/2) I (2x + 2 - 2)/(x^2 + 2x - 2)
=(1/2) I (2x + 2)/(x^2 + 2x - 2) - I 1/(x^2 + 2x - 2)
= (1/2)ln(x^2 + 2x - 2) - I 1/[(x + 1)^2 -3]
=(1/2)ln(x^2 + 2x - 2) - I 1/(x + 1 + sqrt3)(x + 1 - sqrt3)

now just use partial fractions to integrate the socond part.

 

[Arowana21]

So What Do U Get?

 

[ronnknee]

Are we doing your homework for you?

 

[Slidey]

Are we doing your homework for you?

Seems like it.