Integration questionWatch

#1
Hi can anyone help me with this integration question thanks so much

integrate x^2/(1+ x^2) dx
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10 years ago
#2
Substitution?
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10 years ago
#3
Everyone seems to be asking me how to do that integral these days for some reason

Anyway, I think the advice I gave everyone else was to use this substitution:

x = tan(u)

and then you'll end up having to integrate tan^2(u) du

but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)

I think the answer is something like x - arctan(x) + c - Just check that
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10 years ago
#4
.
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#5
(Original post by caaakeeey)
Substitution?
i tried both u = 1 + x^2 and u = x^2 didn't work for me
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10 years ago
#6
When the degree of the numerator is greater than or equal to the degree of the denominator, you want to start by dividing the fraction through (this is essentially what Acid has done).

By degree we mean the highest power of x (and it's only really relevant when dealing with polynomials).
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#7
thanks for your help Glutamic Acid and tazarooni89!
and also caaakeeey
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10 years ago
#8
(Original post by tazarooni89)
Everyone seems to be asking me how to do that integral these days for some reason

Anyway, I think the advice I gave everyone else was to use this substitution:

x = tan(u)

and then you'll end up having to integrate tan^2(u) du

but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)

I think the answer is something like x - arctan(x) + c - Just check that
Yes, I see what you mean

but I think the answer is
x - tan ^-1(x) = x - arctan(x)
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10 years ago
#9
(Original post by Stricof)
Yes, I see what you mean

but I think the answer is x - tan ^-1(x)
Isn't that what I said?

x - arctan(x) + c

c is the constant of integration remember
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10 years ago
#10
(Original post by tazarooni89)
Isn't that what I said?

x - arctan(x) + c

c is the constant of integration remember
Also i edited my answer lol
Sorry man

I got x - arctan(x)
Guess...its' the same LOL!
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#11
(Original post by Swayum)
When the degree of the numerator is greater than or equal to the degree of the denominator, you want to start by dividing the fraction through (this is essentially what Acid has done).

By degree we mean the highest power of x (and it's only really relevant when dealing with polynomials).
got it thanks
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10 years ago
#12
Can't you just use the quotient rule?

(1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

(2x^2x) - (2x^3) / (1+ x^2)^2

2x / (1+ x^2)^2
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10 years ago
#13
(Original post by Judas the Prude)
Can't you just use the quotient rule?

(1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

(2x^2x) - (2x^3) / (1+ x^2)^2

2x / (1+ x^2)^2
No. That's for differentiation.
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10 years ago
#14
(Original post by Judas the Prude)
Can't you just use the quotient rule?

(1+x^2)*(2x) - x^2*2x / (1+ x^2)^2

(2x^2x) - (2x^3) / (1+ x^2)^2

2x / (1+ x^2)^2
Yes, if the question was to differentiate it. But we're integrating here
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10 years ago
#15
(Original post by Glutamic Acid)
No. That's for differentiation.

Oh. LOL. I wondered why no one else had suggested it. Fail.

Thanks.
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10 years ago
#16
(Original post by Glutamic Acid)
No. That's for differentiation.
Lol yes thats true...
for no reason but stupidity I differentiated the equation of the OP

got
(-2*x^3)/(1 + x^2)^2 + (2*x)/(1 + x^2)

I think this is wrong

I know this has no relevance to the actual question, but...just wondering
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10 years ago
#17
(Original post by Stricof)
Lol yes thats true...
for no reason but stupidity I differentiated the equation of the OP

got
(-2*x^3)/(1 + x^2)^2 + (2*x)/(1 + x^2)

I think this is wrong

I know this has no relevance to the actual question, but...just wondering
It's not wrong, but it can be simplified to 2x/(1+x^2)^2

Or if you dont like brackets: 2x/x^4 + 2x^2 + 1

If you wanted to differentiate it easily, I'd suggest looking at post #4 first
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10 years ago
#18
(Original post by tazarooni89)
It's not wrong, but it can be simplified to 2x/(1+x^2)^2
Oh yes...damn, just like the post above my original answer

Okay that was a fail on my accord as well lol
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10 years ago
#19
(Original post by tazarooni89)
Everyone seems to be asking me how to do that integral these days for some reason

Anyway, I think the advice I gave everyone else was to use this substitution:

x = tan(u)

and then you'll end up having to integrate tan^2(u) du

but remember, tan^2(u) = sec^2(u) - 1, and sec^2(u) integrates to give tan (u)

I think the answer is something like x - arctan(x) + c - Just check that
But does it not become 1/(1+x^2) = 1/(1+ tan^2 u) = 1/ sec^2 u = cos^2 u?

Thanks
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10 years ago
#20
But does it not become 1/(1+x^2) = 1/(1+ tan^2 u) = 1/ sec^2 u = cos^2 u?

Thanks
No, because when you use the substitution x = tan u
you also have to remember to say that dx = sec^2(u) du
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