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    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?
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    (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 \int x^2\, dx = \int x\, dx \times \int x\, dx, 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 t=\tan \frac{x}{2}.


    (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 \dfrac{A}{x} + \dfrac{Bx+C}{x^2-2x+5} 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 x = \tan u here.
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    (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 x=\frac{5}{2}\cosh \theta -\frac{1}{2} (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:
    \dfrac{\cos x}{1+\cos x} \equiv \dfrac{1+\cos x}{1+\cos x} - \dfrac{1}{1+\cos x} \equiv 1 - \dfrac{1-\cos x}{1-\cos ^2x} \equiv 1- cosec ^2x + \cot x cosec x

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

    For the other two, nuodai's advice is best.
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    For the first one, you could also redo your integral of \frac{1}{\sqrt{x^2+x-6}}, only this time do it by parts (with du = 1). It's pretty easy to rework the result to get an answer.
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    (Original post by Farhan.Hanif93)
    Try completing the square and using a substitution of x=\frac{5}{2}\cosh \theta (Try to justify why this helps for yourself).
    I'm guessing you meant x+\frac{1}{2} = \frac{5}{2} \cosh \theta here.
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    (Original post by nuodai)
    I'm guessing you meant x+\frac{1}{2} = \frac{5}{2} \cosh \theta here.
    Woops. Yep, that's what I meant. :facepalm:
 
 
 
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