how to do intergration of

Both product and quotient rule are for differentiating a product.  It just happens that the "product" involves a factor in the denominator when using the quotient rule.

In fact, I never bother trying to remember the quotient rule.  I just use the product rule combined with the chain rule to differentiate a quotient.  So, for example I would differentiate y = u/v as follows

y = u*v^-1

dy/dx = u*d(v^-1)/dx + v^-1*du/dx

dy/dx = u*d(v^-1)/dv*dv/dx + v^-1*du/dx

dy/dx = -(uv^-2)*dv/dx + v^-1*du/dx

dy/dx = -(u/v²)*dv/dx +(1/v)*du/dx

Of course this can be rearranged to look like the standard quotient formula, but I find it easier just to apply the steps above directly to the specific functions involved!

wrong and removed

It can be done more simply Stu.

Let y = x² + 1

Then dy = 2xdx

∫2x/(x²+1)dx = ∫dy/y = ln(y) = ln(x² + 1)  (plus an arbitrary constant of integration of course).

Alan would it be a product rule or quotient rule? One is for integral and one for derivatives, opposites to each other?

i was going to go with substitution method haha, i still get confused by making the term dy but a little ok i guess. would the product rule have worked or not worked?

but they cannot solve for integal?

In the case of the integral you use the product form the other way round

Starting with  d(u*v)/dx = u*dv/dx + v*du/dx

Rewrite as  u*dv/dx = d((u*v)/dx - v*du/dx

Integrate each term 

∫udv = u*v - ∫vdu

i saw that on the formulas and thanks.

Stu and Archy

$$\int \frac{f'(x)}{f(x)}dx=ln[f(x)]+c$$

This is an integral that you need to automatically recognise!