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

[FurryLemon]

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.

 

[ShayanJ]

The point is, \ln{ab}=\ln{a}+\ln{b}. Now if you assume c=\frac 1 4 \ln{2} in the second solution, you'll get the first. So they differ only by an additive constant which makes them equivalent.

 

[FurryLemon]

Ok, thanks
Ideally its better to not factor anything out from the denominator then, unless you think you can get partial fractions that is.

 

[STEMucator]

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 $\frac{1}{2}$.

 

[pasmith]

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