# Math Help - Help with integral

1. ## Help with integral

How do I integrate (sqrt(x-2))(x2 + x + 1)

I've got a feeling it's some basic method of integration that I'm forgetting, but I can't find anything like it in my notes.

2. ## Re: Help with integral

substitute :

$u=\sqrt{x-2}$

3. ## Re: Help with integral

I tried that. It doesn't really get me anywhere.

4. ## Re: Help with integral

Originally Posted by necromanzer52
It doesn't really get me anywhere.

5. ## Re: Help with integral

u = rt(x-2)

du/dx = (1/2)(x-2)^(1/2) = 1/(2(rt(x-2)))

(2(rt(x-2))du = dx

Subbing that back in gives me :

u((x2 + x + 1)/((2(rt(x-2))) du

I could cancel out the u to get (1/2)(x2 + x + 1) du, but that's no use.

6. ## Re: Help with integral

I would differentiate it in the form $u^2=x-2$ to get $2udu=dx$

You can also then rearrange to get $x=u^2+2$ then it all comes out nicely when you sub in.

7. ## Re: Help with integral

Originally Posted by necromanzer52
How do I integrate (sqrt(x-2))(x2 + x + 1)

I've got a feeling it's some basic method of integration that I'm forgetting, but I can't find anything like it in my notes.
Start by writing

\displaystyle \begin{align*} t &= x - 2 \\ x &= t + 2 \\ x^2 &= (t + 2)^2 \\ x^2 &= t^2 + 4t + 4 \\ x^2 + x &= t^2 + 4t + 4 + t + 2 \\ x^2 + x &= t^2 + 5t + 6 \\ x^2 + x + 1 &= t^2 + 5t + 7 \\ x^2 + x + 1 &= (x - 2)^2 + 5(x - 2) + 7 \end{align*}

Now rewrite the integrand as \displaystyle \begin{align*} \frac{\sqrt{x - 2}}{x^2 + x + 1} = \frac{2(x - 2)}{2\sqrt{x - 2}\left[(x - 2)^2 + 5(x - 2) + 7\right] } \end{align*}

Now if you make the substitution \displaystyle \begin{align*} u = \sqrt{x - 2} \implies du = \frac{1}{2\sqrt{x - 2}}\,dx \end{align*} the integral becomes

\displaystyle \begin{align*} \int{\frac{2u^2}{u^4 + 5u^2 + 7}\,du} \end{align*}

Can you go from here?

8. ## Re: Help with integral

I can go from there alright. I don't really follow you after you wrote "Now rewrite the integrand as". They're multiplying rather than dividing.

But the other guys method works fine. So I'm grand.