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# Integration watch

1. There are a few which I do not know how to do.

1. int: 1/(x^2+x-6)^3/2 (The previous questions has asked me to find the integral of 1/(x^2+x-6) and the integral of 1/(x^2+x-6)^1/2. Could I just simply multiply my 2 previous answers to obtain the answer to this question?)

2. int: cosx/(1+cosx) (I think I need to use sub. Would u=1+cosx?)

3. int: x^2+2/x(x^2-2x+5) ( I was thinking about partial fractions but I can't simplify the denominator, unless I complete the square. But I have not split a question into partial fractions by completing the square before)

4. int: (1+x^2)^0.5

For 4, I used substitution and let u^2=1+x^2, so 2u.du/dx=2x -> dx=u/x.du
int: u^2.(u/x)du -> 1/x.int u^3 du = 1/x [u^4/4] = (1+x^2)^2/4x+C?
2. (Original post by Superman_Jr)
1. int: 1/(x^2+x-6)^3/2 (The previous questions has asked me to find the integral of 1/(x^2+x-6) and the integral of 1/(x^2+x-6)^1/2. Could I just simply multiply my 2 previous answers to obtain the answer to this question?)
No; does , for example? [It doesn't.] You most likely need to just use a similar method to the cases where the power is 1 or 1/2.

(Original post by Superman_Jr)
2. int: cosx/(1+cosx) (I think I need to use sub. Would u=1+cosx?)
It looks like a t-substitution might be more appropriate here. That is, substitute .

(Original post by Superman_Jr)
3. int: x^2+2/x(x^2-2x+5) ( I was thinking about partial fractions but I can't simplify the denominator, unless I complete the square. But I have not split a question into partial fractions by completing the square before)
You can write this in the form and split it into partial fractions that way. Remember that the denominators in partial fractions don't have to be linear factors, they can be quadratic (or cubic or quartic etc...) -- as long as the numerator has degree one less than the denominator you're fine.

(Original post by Superman_Jr)
4. int: (1+x^2)^0.5

For 4, I used substitution and let u^2=1+x^2, so 2u.du/dx=2x -> dx=u/x.du
int: u^2.(u/x)du -> 1/x.int u^3 du = 1/x [u^4/4] = (1+x^2)^2/4x+C?
You can't just take x out of the integral like that; it's not a constant. By the looks of it, you need to make the substitution here.
3. (Original post by Superman_Jr)
There are a few which I do not know how to do.

1. int: 1/(x^2+x-6)^3/2 (The previous questions has asked me to find the integral of 1/(x^2+x-6) and the integral of 1/(x^2+x-6)^1/2. Could I just simply multiply my 2 previous answers to obtain the answer to this question?)
Try completing the square and using a substitution of (Try to justify why this helps for yourself).

2. int: cosx/(1+cosx) (I think I need to use sub. Would u=1+cosx?)
Alternatively to the t-sub recommended by nuodai, you don't actually need to put as much work as the use of substitution requires - note that:

Where the integrand consists completely on standard integrals if you check out the formula book.

For the other two, nuodai's advice is best.
4. For the first one, you could also redo your integral of , only this time do it by parts (with du = 1). It's pretty easy to rework the result to get an answer.
5. (Original post by Farhan.Hanif93)
Try completing the square and using a substitution of (Try to justify why this helps for yourself).
I'm guessing you meant here.
6. (Original post by nuodai)
I'm guessing you meant here.
Woops. Yep, that's what I meant.

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