Integral of Product: Exploring du/2 in Problem

[Ry122]

In the following problem where does the du/2 come from?
I don't understand why it isn't just du.
http://users.on.net/~rohanlal/partialdif.jpg

 

[danago]

They have made the substitution u = x^2+1, hence the corrosponding differential du is given by:

du = 2x dx

Hence, after rearranging:

dx = du / 2x

Sub this back into the integral, and the x's will cancel out, leaving du/2.

 

[rwisz]

As a calculus student myself, the method we use is slightly different (and I think more commonly taught).

Instead of keeping the 1/2 in the integral, since it is a constant, pull it out to the front of the entire integral, complete the integral, then multiply it by the 1/2 again.

So $$\int x(x^2+1)^{-1} dx$$
$$u=x^2+1, du=2xdx$$ Now at this step, divide both sides by 2 (which is what they originally did in the problem, but instead of just leaving du/2, leave it has $$\frac{1}{2} \int \frac{du}{u}$$

This is a much simpler notation, since you now have a simple du/u to integrate.

Integrate, $$ln(x^2+1)$$ then multiply by 1/2!

$$\frac{1}{2} ln(x^2+1)$$ This also makes the logarithmic operation of raising to the 1/2 easier to spot.

Again, all this is quite trivial as it has no actual effect on the practice of integration. But it IS an easier notation to work with (or so I believe).

 

[Ry122]

when differentiating u why does it gain a dx?
dx is usually only present when attempting to find the integral of something, not when stating the derivative of a function.