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# Integration Q watch

1. Integrate xe^x^2

So I am going to use the rule f'(x)/f(x) ... However it's not a fraction, can I still apply it here?

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2. I'm sorry, I meant the kf'(x)[f(x)]^n rule

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3. But my problem is, we haven't raised the power - [f(x)]^n - during the working out

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4. (Original post by ps1265A)
Integrate xe^x^2

So I am going to use the rule f'(x)/f(x) ... However it's not a fraction, can I still apply it here?

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Just do it by parts

uv - INT[v du/dx]dx
5. (Original post by ps1265A)
Integrate xe^x^2

So I am going to use the rule f'(x)/f(x) ... However it's not a fraction, can I still apply it here?

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Do it by substitution.
6. (Original post by Jackasnacks)
Just do it by parts

uv - INT[v du/dx]dx
Haven't learnt these methods yet

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7. (Original post by ps1265A)
Haven't learnt these methods yet

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Do it by substitution as suggested.

Once you've done it this way, look at the result and ask yourself how you could have seen the answer "by inspection" in the first place
8. (Original post by ps1265A)
Haven't learnt these methods yet

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Use a substitution.
9. You don't need parts for this - you can spot that the 'x' comes from differentiating the power of e.
10. (Original post by ps1265A)
Haven't learnt these methods yet

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Ohh my apologies I misread as xe^2x oops!
Definitely by inspection then
11. (Original post by davros)
Do it by substitution as suggested.

Once you've done it this way, look at the result and ask yourself how you could have seen the answer "by inspection" in the first place
Really struggling to spot do inspection Qs, any tips?

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12. (Original post by ps1265A)
Really struggling to spot do inspection Qs, any tips?

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Use the substitution u = x2.

In the future after practice you will be able to use what I call inverse chain rule in which every step in chain rule is reversed.
13. (Original post by morgan8002)
Use the substitution u = x2.

In the future after practice you will be able to use what I call inverse chain rule in which every step in chain rule is reversed.
Yes, I've learnt it, I just struggle to use inspection for Qs such integrate sec^2xtan^2x

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14. (Original post by ps1265A)
Yes, I've learnt it, I just struggle to use inspection for Qs such integrate sec^2xtan^2x

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If you're struggling on a question like that, do a full substitution and lay all of your working out.
15. (Original post by ps1265A)
I'm sorry, I meant the kf'(x)[f(x)]^n rule

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To the power n is a fairly specific example of kf'(x)g'(f(x)) since this is the chain rule
16. (Original post by TenOfThem)
To the power n is a fairly specific example of kf'(x)g'(f(x)) since this is the chain rule
kf'(x)g'(f(x))

How does this link with the original Q?

Let's say f(x) = e^x^2
f'(x) = 2xe^x^2

The form we could possibly use:
kf'(x)g'(f(x))
Therefore, it wouldn't work. xe^x^2 looks nothing like f(x) multiplied by f'(x)

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17. (Original post by ps1265A)
kf'(x)g'(f(x))

How does this link with the original Q?

Let's say f(x) = e^x^2
f'(x) = 2xe^x^2

The form we could possibly use:
kf'(x)g'(f(x))
Therefore, it wouldn't work. xe^x^2 looks nothing like f(x) multiplied by f'(x)

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That is not how the chain rule works

f(x) = x^2
g(x) = e^x
18. (Original post by TenOfThem)
That is not how the chain rule works

f(x) = x^2
g(x) = e^x

I'm trying to use this here

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19. (Original post by ps1265A)

I'm trying to use this here

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But that douse not work here as you do not have g(x) in the form x^(n+1)
20. (Original post by TenOfThem)
But that douse not work here as you do not have g(x) in the form x^(n+1)
That's why I'm confused. The exercise I'm doing ONLY concerns the 2 methods I mentioned right at the beginning of the post.

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