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# P2 / P3 question of Integration watch

1. Im doing P3, P1 and S1 next week.

How do you integrate 8/ x^2 + 1 ????

Can you integrate where the denominator has a power???

2. Is that 8 / (x^2 + 1) or (8 / x^2) + 1 ?
3. (Original post by BlueAngel)
Im doing P3, P1 and S1 next week.

How do you integrate 8/ x^2 + 1 ????

Can you integrate where the denominator has a power???

you need to specify what is the denominator.

i suspect its (x^2 + 1) thats the total denominator (if it was just x^2, thats p1)

there are several ways to do it, but the easiest is to look in the formula book other wise, some sort of substitution will do it.
4. (Original post by Morgan)
Is that 8 / (x^2 + 1) or (8 / x^2) + 1 ?
Its 8 / (x^2 + 1)

Actually this is more difficult, its in the P3 exam paper.

Evaluate this Limits are x = square root of 2 and x = 0

Integral of x / squareroot of (16 - x^4) -

Ive got the answers, just doesnt explain it.
5. (Original post by BlueAngel)
Im doing P3, P1 and S1 next week.

How do you integrate 8/ x^2 + 1 ????

Can you integrate where the denominator has a power???

Let x=tanu
dx=sec^2u.du

int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.
6. (Original post by BlueAngel)
Its 8 / (x^2 + 1)

Actually this is more difficult, its in the P3 exam paper.

Evaluate this Limits are x = square root of 2 and x = 0

Integral of x / squareroot of (16 - x^4) -

Ive got the answers, just doesnt explain it.
Okay you can just substitute and use u = x^2 +1 I think and glancing at the second one I think its a nominator-derivative of the denominator one.
7. (Original post by Ralfskini)
Let x=tanu
dx=sec^2u.du

int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.

OK RALFSKINI PLEASE TELL ME WHY YOU WOULD USE TAN HAS A SUBSTITUTION WHEN THE EQUATION INVOLVES ONLY NUMBERS, COS I AM SO CONFUSED NOW
8. (Original post by BlueAngel)
Im doing P3, P1 and S1 next week.

How do you integrate 8/ x^2 + 1 ????

Can you integrate where the denominator has a power???

The derivtative of tan^-1(x) is a standard form and is 1/(x^2 + 1).

Using this, it's simply 8tan^-1x + C.

The standard form is INT c/(ax^2 + b^2) = c.tan^-1(x/b)
9. (Original post by calumc)
The derivtative of tan^-1(x) is a standard form and is 1/(x^2 + 1).

Using this, it's simply 8tan^-1x + C.

The standard form is INT c/(ax^2 + b^2) = c.tan^-1(x/b)

Is this from P3??
10. (Original post by BlueAngel)
OK RALFSKINI PLEASE TELL ME WHY YOU WOULD USE TAN HAS A SUBSTITUTION WHEN THE EQUATION INVOLVES ONLY NUMBERS, COS I AM SO CONFUSED NOW

You can make this substitution because of the trig identity- 1+tan^x=sec^2x.
11. how do u guess wot to use for u in subtitution?
12. (Original post by BlueAngel)
Is this from P3??
I have no idea, I did Advanced Highers which are slightly different, and harder apparently.
13. (Original post by Ralfskini)
Let x=tanu
dx=sec^2u.du

int.8/x^2+1.dx=8int.sec^2u/sec^2u.du=8u+C=8arctanx+C.
being able to get arctan from an integration isn't on P3!!!

Can someone please tell me how to do this question - cos it's getting me worried
14. (Original post by Silly Sally)
being able to get arctan from an integration isn't on P3!!!

Can someone please tell me how to do this question - cos it's getting me worried

..but I have only done up to P3.
15. (Original post by Ralfskini)
..but I have only done up to P3.
Do you do edexcel or another board?
16. (Original post by Silly Sally)
Do you do edexcel or another board?

Edexcel.
17. (Original post by Ralfskini)
Edexcel.
AGHHHHH!! Same as me!!!! I can't do this integration. I would at first have involved an 'ln', like integral of 1/(3x + 1) = 1/3 ln (3x +1) <--- i think - now i don't know whther this is right or wrong!!!
18. (Original post by Silly Sally)
AGHHHHH!! Same as me!!!! I can't do this integration. I would at first have involved an 'ln', like integral of 1/(3x + 1) = 1/3 ln (3x +1) <--- i think - now i don't know whther this is right or wrong!!!

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