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Homework Help: When Integrating (2x)/(4x^(2)+2) I get two different integrals ?

  1. Aug 19, 2014 #1

    So lets have ∫(2x)/(4x^(2)+2) dx

    Without factorising the 2 from the denominator, I integrate and I get

    1/4*ln(4x^(2)+2)+c which makes sense as when I differentiate it I get the original derivative.

    BUT when I factor the 2 from the denominator I have

    2x/[2(2x^(2)+1)] simplify it down = x/(2x^(2)+1) which is the same as (2x)/(4x^(2)+2)

    Now when I integrate ∫x/(2x^(2)+1)dx I get 1/4*ln(2x^(2)+1)+c

    which is obviously different from the first integral. Why? because when I simplify it by factorisation its the same thing so Why is it different?

    Please, any help would be appreciated.

  2. jcsd
  3. Aug 19, 2014 #2


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    The point is, [itex] \ln{ab}=\ln{a}+\ln{b} [/itex]. Now if you assume [itex] c=\frac 1 4 \ln{2} [/itex] in the second solution, you'll get the first. So they differ only by an additive constant which makes them equivalent.
    Last edited: Aug 19, 2014
  4. Aug 19, 2014 #3
    Ok, thanks
    Ideally its better to not factor anything out from the denominator then, unless you think you can get partial fractions that is.
    Last edited: Aug 19, 2014
  5. Aug 19, 2014 #4


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    Indeed, this can be observed from something as simple as:

    $$\int x-1 \space dx$$

    Making a substitution ##u = x - 1## gives ##du = dx##. Hence:

    $$\int x-1 \space dx = \frac{(x-1)^2}{2} + C$$

    Now what if you simply integrated by using some integral properties instead of a substitution?

    $$\int x-1 \space dx = \int x \space dx - \int 1 \space dx = \frac{x^2}{2} - x + K$$

    The two answers look nothing alike, but they're exactly the same! In fact, you can show they differ by a constant. Taking the answer for the substitution method and massaging it a bit gives you:

    $$\frac{(x-1)^2}{2} + C = \frac{x^2 - 2x + 1}{2} + C = \frac{x^2}{2} - x + \frac{1}{2} + C$$

    It is clear by observation with the non substitution answer that ##K = \frac{1}{2} + C## so ##K## and ##C## differ by ##\frac{1}{2}##.
  6. Aug 19, 2014 #5


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    The other classic example is [tex]\int \cos \theta \sin\theta\,d\theta = -\frac12\cos^2\theta + C = \frac12\sin^2\theta - \frac12 + C = -\frac14 \cos(2\theta) + C - \frac14.[/tex]
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