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Intergration help

What method of intergration can I use if I cannot factorise the denominator, so cannot use partial fractions.
i.e intergrate x+1/x^2 + 2x + 2

Any help would be greatly appreciated :smile:

Even just like a nudge in the right direction would be helpful
(edited 12 years ago)
Reply 1
Okay so the derivative of the denominator= 2x+2 right but you have x+1 in the denominator.

Pull a half outside the integral and put 2x+2 as the numerator and you have 1/2 * 2x+2/(x^2+2x+2)

Now you have the integral of f'(x)/f(x). Im sure you can do the same.

Why the neg? My advice is correct!
(edited 12 years ago)
Reply 2
Let u=x2+2x+2\text{Let }u=x^2+2x+2

If you can't do it even with the substitution I'll show you the method :smile:
Reply 3
Try taking 1/2 out, so it is 1/2 integral of (2x + 2)/(x^2 + 2x + 2). Does that look like anything familiar?

Spoiler

Reply 4
Original post by starkrush
Try taking 1/2 out, so it is 1/2 integral of (2x + 2)/(x^2 + 2x + 2). Does that look like anything familiar?

Spoiler



Thank you for the reply, so the right answer would be 1/2 ln f(x) + c right
Reply 5
Original post by f1mad
Okay so the derivative of the denominator= 2x+2 right but you have x+1 in the denominator.

Pull a half outside the integral and put 2x+2 as the numerator and you have 1/2 * 2x+2/(x^2+2x+2)

Now you have the integral of f'(x)/f(x). Im sure you can do the same.


Thank you, im not sure why anyone negged your response :/
would the correct answer then be 1/2 ln f(x) + c
Reply 6
Your answer is correct yeah
Reply 7
Original post by anonstudent1
Thank you, im not sure why anyone negged your response :/
would the correct answer then be 1/2 ln f(x) + c


Yeah we really should have the facility to see who negs others for no apparent reason. Yes it would be correct.
Reply 8
Original post by f1mad
Yeah we really should have the facility to see who negs others for no apparent reason. Yes it would be correct.


Il pos rep you tomorrow, even though mines worthless :biggrin:

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