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

• Nov 13th 2011, 01:26 AM
angypangy
root of x in denominator and not yet been taught to integrate quotients
Hello

$\displaystyle \int \frac{3x+1}{\sqrt(2)}$

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
• Nov 13th 2011, 01:29 AM
Siron
Re: root of x in denominator and not yet been taught to integrate quotients
You can put the constant $\displaystyle \frac{1}{\sqrt{2}}$ outside the integral:
$\displaystyle \int \frac{3x+1}{\sqrt{2}}dx=\frac{1}{\sqrt{2}}\int (3x+1)dx$

Now you have a basic integral.
• Nov 13th 2011, 01:55 AM
e^(i*pi)
Re: root of x in denominator and not yet been taught to integrate quotients
Quote:

Originally Posted by angypangy
Hello

$\displaystyle \int \frac{3x+1}{\sqrt(2)}$

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

$\displaystyle \sqrt{2}$ is a constant and so therefore is it's reciprocal $\displaystyle \dfrac{1}{\sqrt{2}}$. If you've been taught how to rationalise the denominator then this is also $\displaystyle \dfrac{\sqrt2}{2}$ which is still a constant.

$\displaystyle \int \dfrac{3x+1}{\sqrt{2} dx} = \dfrac{\sqrt2}{2}\int (3x+1) dx$

Note that $\displaystyle \dfrac{\sqrt{2}}{2}C$ is also a constant (where C is your normal constant of integration)
• Nov 13th 2011, 02:29 AM
Prove It
Re: root of x in denominator and not yet been taught to integrate quotients
Quote:

Originally Posted by angypangy
Hello

$\displaystyle \int \frac{3x+1}{\sqrt(2)}$

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: $\displaystyle \displaystyle \int{\frac{3x + 1}{\sqrt{x}}\,dx}$, in which case...

\displaystyle \displaystyle \begin{align*} \int{\frac{3x+1}{\sqrt{x}}\,dx} &= \int{\frac{3x}{\sqrt{x}} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3\sqrt{x} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3x^{\frac{1}{2}} + x^{-\frac{1}{2}}\,dx} \end{align*}

and you should now be able to integrate using the rules you know :)
• Nov 15th 2011, 05:48 AM
angypangy
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: $\displaystyle \displaystyle \int{\frac{3x + 1}{\sqrt{x}}\,dx}$, in which case...

\displaystyle \displaystyle \begin{align*} \int{\frac{3x+1}{\sqrt{x}}\,dx} &= \int{\frac{3x}{\sqrt{x}} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3\sqrt{x} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3x^{\frac{1}{2}} + x^{-\frac{1}{2}}\,dx} \end{align*}

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.