# Homework Help: Integral Help Please

1. May 3, 2004

### Zurtex

Got an integral that for some reason I can't do, very much annoying me because this is supposed to be the stuff I am good at. A nudge in the right direction would be great, here it is:

$$\int \frac{(x+2)dx}{\sqrt{x^2+6x+4}}$$

I have attempted completing the square on the bottom to get $(x+3)^2-5$ and then I tried the substitution $u=x+3$. After messing about with it a bit I got an answer but it didn't look right and it was a lot different from the answer I got off an online integrator and its answer really did look better.

Any help would be great thanks.

2. May 3, 2004

### arildno

Have you tried the substitution:
u= sqrt(5)*Cosh(t)?

3. May 3, 2004

### Zurtex

Thanks, could someone please confirm this is the answer then:

$$\sqrt{ \frac{x^2-6x+4}{5} } - \cosh^{-1}\left(\frac{x-3}{\sqrt{5}}\right) + C$$

4. May 3, 2004

### arildno

Well, I got the -6x term to be +6x, and the x-3 to be x+3

5. May 3, 2004

### HallsofIvy

Actually, I like your first idea of substituting for x+3. If you let u= x+3, then
x2+6x+ 4= (x+3)2- 5= u2- 5 and x+2= x+3-1= u-1 so the integral becomes
$$\int{\frac{u-1}{\sqrt{u^2-5}}du$$
which we can separate into
$$\int{\frac{udu}{\sqrt{u^2-5}}+ \int{\frac{du}{\sqrt{u^2-5}}$$

Now you can make the substitution v= u2-5 in the first integral and
$$sec(\theta)= \sqrt{5}u$$ in the second.

Last edited by a moderator: May 4, 2004
6. May 3, 2004

### Zurtex

kk thanks for the help and thanks for the other approach

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