# Math Help - Integrate a fraction with limits, with the route of x on the bottom

1. ## Integrate a fraction with limits, with the route of x on the bottom

I have no idea how to do this:

Evaluate the following integral::

$\int \frac {1}{\sqrt{x+1}} dx$

I thought the bottom can be separated into x^-.05 and 1 which wouldn't be too bad to go from there but I don't think that's the answer.

As soon as I can get x to a power I think I can integrate it into ln(something) but I'm not sure how to do that first step.

2. ## Re: Integrate a fraction with limits, with the route of x on the bottom

Originally Posted by Mukilab
I have no idea how to do this:
Evaluate the following integral::
$\int \frac {1}{\sqrt{x+1}} dx$
What is the derivative of $2\sqrt{x+1}~?$

3. ## Re: Integrate a fraction with limits, with the route of x on the bottom

oh like chain rule?

So the derivative of that is

2(0.5)(1)(x+1)^{-0.5} = (x+1)^{-0.5}

Huh. How did you figure out it's 2sqrt(x+1) though? Are there any steps or was it just intuition?

4. ## Re: Integrate a fraction with limits, with the route of x on the bottom

Originally Posted by Mukilab
I have no idea how to do this:

Evaluate the following integral::

$\int \frac {1}{\sqrt{x+1}} dx$

I thought the bottom can be separated into x^-.05 and 1 which wouldn't be too bad to go from there but I don't think that's the answer.

As soon as I can get x to a power I think I can integrate it into ln(something) but I'm not sure how to do that first step.
\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{x + 1}}\,dx} &= \int{ \left( x + 1 \right) ^{-\frac{1}{2}}\,dx} \end{align*}

If you're good with your integration processes, you can get the integral from here, otherwise make a substitution \displaystyle \begin{align*} u = x + 1 \implies du = dx \end{align*}.

5. ## Re: Integrate a fraction with limits, with the route of x on the bottom

Thanks.

I have no idea how to integrate something in brackets, so I've gone with the u thing.

So $\int (x+1)^\frac{-1}{2} dx = \int u^\frac{-1}{2} du = [2u^\frac{1}{2}] + c = 2(x+1)^\frac{1}{2} + c = 2 \sqrt{(x+1)} + c$ is that correct?

6. ## Re: Integrate a fraction with limits, with the route of x on the bottom

Yes that's correct, well done