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

In summary: This time, the substitution method yields:$$\int \cos \theta \sin\theta\,d\theta = -\frac12\cos^2\theta + C$$and the non substitution method yields:$$\int \cos \theta \sin\theta\,d\theta = -\frac12\cos^2\theta$$
  • #1
FurryLemon
2
0
Hi

So let's 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.

Thanks.
 
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  • #2
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.
 
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  • #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.
 
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  • #4
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}##.
 
  • #5
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]
 

1. Why do I get two different integrals when integrating (2x)/(4x^2+2)?

When integrating a function, it is possible to get different integrals depending on the method used. In this case, the two different integrals are a result of using two different techniques - integration by parts and substitution. Both methods are valid and can be used to solve the integral.

2. Which method should I use to integrate (2x)/(4x^2+2)?

There is no definitive answer to this question as both integration by parts and substitution can be used to solve the integral. It is often a matter of personal preference or convenience. It is recommended to try both methods and see which one gives a simpler or more manageable solution.

3. Can I choose which integral to use when integrating (2x)/(4x^2+2)?

Yes, you can choose which integral to use when integrating (2x)/(4x^2+2), but it is important to make sure that the chosen method is appropriate for the given integral. Using the wrong method can result in an incorrect solution.

4. Why does one integral result in a logarithmic function while the other results in an arctangent function?

This is a result of the two different methods used. Integration by parts often results in a logarithmic function, while substitution often results in an arctangent function. Both functions are valid solutions to the integral and can be simplified or manipulated further if needed.

5. Are there any other methods I can use to integrate (2x)/(4x^2+2)?

Yes, there are other methods that can be used to integrate (2x)/(4x^2+2), such as partial fractions or trigonometric substitution. It is always helpful to know and understand multiple integration techniques to have a wider range of options when solving integrals.

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