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# integration of 1/(x^2 + 9) watch

1. What's the way to to go about it?

the question was to find the integration of (x^2 + 5) / (x^2 + 9). Dividing the numer. by denom. gives integration of (1 - 4/(x^2 + 9)). Now what?
2. Make the substitution and use the identity .

[Or, if you're lazy, look in the formula book.]
3. so you've got 1-(4/x^2 + 9)
integrating 1 gets you x, so now you have -4/(x^2 + 9)
to integrate
treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
then times it by the -4, to get -2 ln(x^2 + 9)

so overall you get x - 2 ln(x^2 + 9)

sorry if this isn't very clear
4. (Original post by proud nd luvin it)
so you've got 1-(4/x^2 + 9)
integrating 1 gets you x, so now you have -4/(x^2 + 9)
to integrate
treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
then times it by the -4, to get -2 ln(x^2 + 9)

so overall you get x - 2 ln(x^2 + 9)

sorry if this isn't very clear
Isn't integ. of (1/something) = (ln something) just for linear expressions?
5. (Original post by proud nd luvin it)
so you've got 1-(4/x^2 + 9)
integrating 1 gets you x, so now you have -4/(x^2 + 9)
to integrate
treat it as 1/(x^2 + 9) this is (1/2) ln(x^2 + 9)
then times it by the -4, to get -2 ln(x^2 + 9)

so overall you get x - 2 ln(x^2 + 9)

sorry if this isn't very clear
int [f'(x)/f(x)] = ln(f(x))

what you have said here is clearly wrong.

use the tan substitution OP
6. (Original post by nuodai)
Make the substitution and use the identity .

[Or, if you're lazy, look in the formula book.]
But why would x be equal to 3tan(theta)? Because the final answer is given in terms of x
.
7. (Original post by gunmetalpanda)
Isn't integ. of (1/something) = (ln something) just for linear expressions?

(Original post by slylion1)
int [f'(x)/f(x)] = ln(f(x))

what you have said here is clearly wrong.

use the tan substitution OP
Oh crap... yes you're right sorry but still... how come you'd use the tan substitution??
8. (Original post by gunmetalpanda)
But why would x be equal to 3tan(theta)? Because the final answer is given in terms of x
.
You substitute , integrate w.r.t. , and then make the reverse substitution to put your answer in terms of again. You substitute because then the 3 squares to a 9, which you can take out as a common factor, and then you can apply the identity on the denominator of the fraction. Because you get , the terms end up cancelling and you get something simple to integrate.
9. Also, could i just say that
Spoiler:
Show

So in this case, a would be 3.

_Kar.
10. (Original post by Kareir)
Also, could i just say that

Spoiler:
Show

So in this case, a would be 3.

_Kar.
Yes (but that's cheating [I mean not actually cheating, but it takes the fun out of integration]).
11. Haha, i've spoilered it, just to keep a bit of the mystery in

_Kar.

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