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    Hi guys, can someone please check if I am integrating these questions correctly using:Name:  7.PNG
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    For the first problem, the derivative of the bottom isn't equal to the top (or a linear multiple of). Partial fractions seems to be the way into it.
    For the second problem, you made a little mistake.
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    Q1: Was the top line in the question supposed to be 6x + 3x^2 by any chance? If it was, then you have f'(x) / f(x) as per your attachment (5). When integrating with respect to x, you can't take x-terms outside the integral, by the way.

    Q2: your answer looks OK to me.

    General: I would strongly encourage you to check your working in questions like this by differentiating you answer. In fact trial and error can be a good way of establishing the correct multiplying factor.
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    (Original post by thekidwhogames)
    For the first problem, the derivative of the bottom isn't equal to the top (or a linear multiple of). Partial fractions seems to be the way into it.
    For the second problem, you made a little mistake.
    Thanks very much for your reply!
    For the first question:
    What you mentioned about the derivative of the denominator being equal to the numerator or a linear multiple of the numerator means you a cannot do Name:  7.PNG
Views: 15
Size:  34.3 KB if the derivative of the denominator is not equal or a linear multiple of the numerator, right? So you do partial fractions.. I did that and I got an answer of (1/3)ln|x|/|x+3| - 2/x +c ..is that correct?

    Could you please kindly point out where I made my mistake in the second question?
    In question 2, I am confused whether the answer to question 2 is 1/3ln(3x^3+3x^2) +c or 1/3ln(x^3+x^2)+c? Because for question 2 you can decide not to cancel the 3 from the numerator with the 3 from the denominator and so end up with 1/3ln(3x^3+3x^2) +c .

    I am following the same technique I used to integrate 1/4x to integrate question 2. I have attached that here.
    Could you please check I am doing this the right way? Many thanks
    Attachment 723826723828
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    (Original post by old_engineer)
    Q1: Was the top line in the question supposed to be 6x + 3x^2 by any chance? If it was, then you have f'(x) / f(x) as per your attachment (5). When integrating with respect to x, you can't take x-terms outside the integral, by the way.

    Q2: your answer looks OK to me.

    General: I would strongly encourage you to check your working in questions like this by differentiating you answer. In fact trial and error can be a good way of establishing the correct multiplying factor.
    Thanks so much for your reply!!
    In question 2, I am still confused though whether the answer to question 2 is 1/3ln(3x^3+3x^2) +c or 1/3ln(x^3+x^2)+c?
    Because for question 2 you can decide not to cancel the 3 from the numerator with the 3 from the denominator and so end up with 1/3ln(3x^3+3x^2) +c .

    I am following the same technique I used to integrate 1/4x to integrate question 2. I have attached that here.
    Could you please check I am doing this the right way? Many thanks
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    (Original post by sienna2266)
    Thanks so much for your reply!!
    In question 2, I am still confused though whether the answer to question 2 is 1/3ln(3x^3+3x^2) +c or 1/3ln(x^3+x^2)+c?
    Because for question 2 you can decide not to cancel the 3 from the numerator with the 3 from the denominator and so end up with 1/3ln(3x^3+3x^2) +c .

    I am following the same technique I used to integrate 1/4x to integrate question 2. I have attached that here.
    Could you please check I am doing this the right way? Many thanks
    Both of your candidate answers to Q2 are actually correct, and this usefully illustrates a property of logs. Here's how.

    ln(3x^3 + 3x^2) = ln(3(x^3 + x^2)) = ln 3 + ln(x^3 + x^2)

    ln3 is a constant and hence disappears on differentiation. Thus differentiating ln(3x^3 + 3x^2) will give you the same answer as differentiating ln(x^3 + x^2). Try it if not convinced.

    I can't really comment on your method as I'm struggling to follow it, but it seems to work (and you seem to have mastered it), so stick with it, especially if that's what you've been taught.

    As mentioned before, I advocate the "trial" method, which consists of spotting that the top line looks like a multiple of the differential of the bottom line, trying out ln(bottom line) as a candidate solution and inferring the multiplication factor from that.
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    (Original post by old_engineer)
    Both of your candidate answers to Q2 are actually correct, and this usefully illustrates a property of logs. Here's how.

    ln(3x^3 + 3x^2) = ln(3(x^3 + x^2)) = ln 3 + ln(x^3 + x^2)

    ln3 is a constant and hence disappears on differentiation. Thus differentiating ln(3x^3 + 3x^2) will give you the same answer as differentiating ln(x^3 + x^2). Try it if not convinced.

    I can't really comment on your method as I'm struggling to follow it, but it seems to work (and you seem to have mastered it), so stick with it, especially if that's what you've been taught.

    As mentioned before, I advocate the "trial" method, which consists of spotting that the top line looks like a multiple of the differential of the bottom line, trying out ln(bottom line) as a candidate solution and inferring the multiplication factor from that.
    Thanks for your reply back! It cleared up lots of things.
    However,if you sub in for example, x=2 into 1/3ln(3x^3+3x^2)+c you get 1.19+c but if you sub in x=2 into 1/3ln(x^3+x^2)+c you get 0.83+c . But then, the c value is unknown and probably different for 1/3ln(3x^3+3x^2)+c and 1/3ln(x^3+x^2)+c - so subbing in x=2 into both 1/3ln(3x^3+3x^2)+c and 1/3ln(x^3+x^2)+c would give the same f(x) value, right?
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    (Original post by sienna2266)
    Thanks for your reply back! It cleared up lots of things.
    However,if you sub in for example, x=2 into 1/3ln(3x^3+3x^2)+c you get 1.19+c but if you sub in x=2 into 1/3ln(x^3+x^2)+c you get 0.83+c . But then, the c value is unknown and probably different for 1/3ln(3x^3+3x^2)+c and 1/3ln(x^3+x^2)+c - so subbing in x=2 into both 1/3ln(3x^3+3x^2)+c and 1/3ln(x^3+x^2)+c would give the same f(x) value, right?
    That's a good observation. As you've noted, (1/3)ln(3x^3+3x^2) is not the same as (1/3)ln(x^3+x^2). The difference between the two is equivalent to (1/3)ln(3), and this is true for all values of x. The constant c can take any value, so you can think of the constant ln(3) as simply being absorbed into c.

    The functions both have the same gradient (differential) for all values of x, and this is what is important in this context where they are paired with an integral.
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    (Original post by old_engineer)
    That's a good observation. As you've noted, (1/3)ln(3x^3+3x^2) is not the same as (1/3)ln(x^3+x^2). The difference between the two is equivalent to (1/3)ln(3), and this is true for all values of x. The constant c can take any value, so you can think of the constant ln(3) as simply being absorbed into c.

    The functions both have the same gradient (differential) for all values of x, and this is what is important in this context where they are paired with an integral.
    Thanks so much!!

    So if c=0 for y=(1/3)ln(3x^3+3x^2) +c to give y=(1/3)ln(3x^3+3x^2)
    and if c=(1/3)ln3 for y=(1/3)ln(x^3+x^2) +c to give y=(1/3)ln(x^3+x^2) +(1/3)ln3 ...Subbing x=2 would give the same y value of 1.19 for both equations.

    Also, dy/dx is the same for both equations since the c disappears when you differentiate.

    For both equations, dy/dx = (3x^2 +2x)/(x^3 +x^2)
 
 
 
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