could someone please integrate this

[M-D]

for a question like this: integrate 7/(x^2-8x+25)dx, i would use partial fractions but the denominator can not be factorized in this case, therefore i do not know how to do this. please help.

[Alwin]

for a question like this: integrate 7/(x^2-8x+25)dx, i would use partial fractions but the denominator can not be factorized in this case, therefore i do not know how to do this. please help.

hint: complete the square in the denominator. you should get something familiar looking a square plus another square

[Quantum.Mechanic]

Basically complete the square and then you integrate it so that it becomes and an the integrand is inverse tan
So, when you complete the square it becomes
∫▒7/(9+〖(x-4)〗^2 )
Then we let (x-4)=u
Then we factorise it as
7∫1/(u^2+9)
Then we take the 9 out, so that it becomes
7/9∫1/(u^2/9+1)
Therefore ∫1/(1+u^2/9), substitute s = u/3 and  ds = 1/3  du:
This becomes 7/3  ∫1/(1+s^2) ds
= 7/3 tan^(-1)(s)+

And then sub in s = u/3 and u = x-4
 7/3 cotan(1/3 (-4+x))

[Quantum.Mechanic]

Sorry its messy as.
I wish I had a math program.
But I don't know any!

[M-D]

thank you for your help. i understood the whole thing except for the part where 7/9 changed to 7/3, how did you do that?

[Alwin]

Sorry its messy as.
I wish I had a math program.
But I don't know any!

hmm. it  looks to me like u used the equation editor in word. try LaTeX, it's the oi button under the Bold button. there are some explainations and tutorials floating around on this forum

[Quantum.Mechanic]

Here.
Ill clarify it
7/9∫1/(u^2/9+1)
Therefore ∫1/(1+u^2/9).du, substitute s = u/3 and  ds = 1/3  du:
This becomes 7/3  ∫1/(1+s^2) ds
= 7/3 tan^(-1)(s)

Because for this bit, where is ∫1/(1+u^2/9).du, we want it in terms of du. so du = 3.ds
This means that we multiply the whole 7/9 x 3 which would give the 7/3

hmm. it  looks to me like u used the equation editor in word. try LaTeX, it's the oi button under the Bold button. there are some explainations and tutorials floating around on this forum

Ill give this a look.
I've been trying to figure it out since I've been wanting to write up a further mathematics shortcut book.
Ill work it out on the holidays when there is less stress, because I have a crazy amount of hours for my double degree!!!

[M-D]

thanks. i wish you the best of luck in your degree