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How to evaluate definite integral if the limits are 0 to infinity? Watch

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    The indefinite integral is  ln \frac{(1+2x)^2}{1+x} + C , but what is the value of  [ln \frac{(1+2x)^2}{1+x} + C]_0^\infty .

    On one of the example sheets,  [ln \frac{(1+2x)}{1+x} + C]_0^\infty = ln(2); but how?
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    (Original post by Alpha-Omega)

    The indefinite integral is  ln \frac{(1+2x)^2}{1+x} + C , but what is the value of  [ln \frac{(1+2x)^2}{1+x} + C]_0^\infty .
    There shouldn't be a square there - you've made a slip in the integration or the partial fractions.

    On one of the example sheets,  [ln \frac{(1+2x)}{1+x} + C]_0^\infty = ln(2); but how?
    Well the C will cancel, and the lower limit is otherwise 0.

    For the upper limit, take the limit as x goes to infinity. It would help to rewrite (1+2x)/(1+x) as 2 - 1/(1+x), in which case the limit should be clear.
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    (Original post by ghostwalker)
    There shouldn't be a square there - you've made a slip in the integration or the partial fractions.



    Well the C will cancel, and the lower limit is otherwise 0.

    For the upper limit, take the limit as x goes to infinity. It would help to rewrite (1+2x)/(1+x) as 2 - 1/(1+x), in which case the limit should be clear.
    That C - must be the coffee.

    I was wondering about that too, the answer sheet does not have square too, as you say.

    I get the following for the partial fraction:
    \frac{1}{(1+x)(1+2x)} = \frac{2}{1+2x} - \frac{1}{1+x}

    So, shouldn't the integral be:

    2ln(1+2x) - ln(1+x), and

    \lambda ln(x) = ln(x^\lambda),

    so it must be 2ln(1+2x) = ln(1+2x)^2?
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    Apart from that, I understand now how ln(2) is achieved.
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    (Original post by Alpha-Omega)
    That C - must be the coffee.

    I was wondering about that too, the answer sheet does not have square too, as you say.

    I get the following for the partial fraction:
    \frac{1}{(1+x)(1+2x)} = \frac{2}{1+2x} - \frac{1}{1+x}

    So, shouldn't the integral be:

    2ln(1+2x) - ln(1+x), and

    \lambda ln(x) = ln(x^\lambda),

    so it must be 2ln(1+2x) = ln(1+2x)^2?
    You don't need the 2 in front - remember that d/dx(1+2x) = 2.

    To evaluate the limits, decompose it into 1/(1+x) + 2x/(1+x).

    As x tends to infinity, 1/(1+x) clearly tends to 0. Now, what happens to 2x/(1+x) as x tends to infinity? Think of how significant (1+x) is compared to x as x gets big (say, 1,000,001 compared to 1,000,000).

    As x tends to 0, 1/(1+x) clearly tends to 1, and 2x/(1+x) also has an obvious limit.

    What happens when you combine all of these things?
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    (Original post by Hedgeman49)
    You don't need the 2 in front - remember that d/dx(1+2x) = 2.

    To evaluate the limits, decompose it into 1/(1+x) + 2x/(1+x).

    As x tends to infinity, 1/(1+x) clearly tends to 0. Now, what happens to 2x/(1+x) as x tends to infinity? Think of how significant (1+x) is compared to x as x gets big (say, 1,000,001 compared to 1,000,000).

    As x tends to 0, 1/(1+x) clearly tends to 1, and 2x/(1+x) also has an obvious limit.

    What happens when you combine all of these things?
    Ok, I get it now.
 
 
 
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