root of x in denominator and not yet been taught to integrate quotients

Hello

And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

Angus

Re: root of x in denominator and not yet been taught to integrate quotients

You can put the constant outside the integral:

Now you have a basic integral.

Re: root of x in denominator and not yet been taught to integrate quotients

Quote:

Originally Posted by

**angypangy** Hello

And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

Angus

is a constant and so therefore is it's reciprocal . If you've been taught how to rationalise the denominator then this is also which is still a constant.

Note that is also a constant (where C is your normal constant of integration)

Re: root of x in denominator and not yet been taught to integrate quotients

Quote:

Originally Posted by

**angypangy** Hello

And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

Angus

I expect, from the title of the thread, that the OP meant to ask how to integrate this: , in which case...

and you should now be able to integrate using the rules you know :)

Re: root of x in denominator and not yet been taught to integrate quotients

Quote:

Originally Posted by

**Prove It** I expect, from the title of the thread, that the OP meant to ask how to integrate this:

, in which case...

and you should now be able to integrate using the rules you know :)

Yes this was the question - but glad I got it wrong - because answers to my incorrect question were useful. I had a think a think just after posting and worked out what to do. But thanks for replying.