# Intergal of x^2/sqrt(x^2+3)

1. Sep 6, 2011

### ParoXsitiC

1. The problem statement, all variables and given/known data

$\int \frac{x^2}{\sqrt{x^2+3}}$

2. Relevant equations

sinh-1(u) = u' / (u^2 + 1)

3. The attempt at a solution

Make the x^2 + 3 look like x^2 + 1 by taking out a sqrt(3). Giving you

$\int \frac{x^2}{\sqrt{3} \sqrt{\frac{x^2}{3}+1}}$

Set the constant outside the integral.

$\frac{1}{\sqrt{3}} \int \frac{x^2}{\sqrt{\frac{x^2}{3}+1}}$

Now we find where $u^2 = \frac{x^2}{3}$ , which is $u = \frac{x}{\sqrt{3}}$. Now we know the u of the sinh-1, we find u'

$u' = \frac{1}{\sqrt{3}}$

So now we taken care of everything but x^2...

Where to go now?

2. Sep 6, 2011

### micromass

So, now substite in the sinh. You'll see that it works out fine...

What do you get??

3. Sep 6, 2011

### ParoXsitiC

I know that the sinh^-1(x/sqrt(3)) is in the answer, but there is still a multiplication between x^2 and the the sin so more needs to be done.

I can't just have

(1/3)x^3 * sinh^-1(x/sqrt(3)) + c

because the product rule states uv' + u'v, if it was as easy as u'v' then it would work.

4. Sep 6, 2011

### micromass

You have the integral

$$\int{\frac{x^2}{\sqrt{x^2+1}}dx}$$

after your substitutions right?? So just substitute in $x=sinh(u)$

5. Sep 6, 2011

### HallsofIvy

Micromass likes his hyperbolic functions substitutions! Perhaps just because i learned them first, I always think of trig substitutions first. Here, after taking out a $\sqrt{3}$, you have $\sqrt{1+(x/\sqrt{3})^2}$ and since $1+ tan^2(\theta)= sec^2(\theta)$ I would let $x= \sqrt{3}tan(\theta)$.

Of course, then, the $x^2$ in the numerator becomes $3tan^2(\theta)$

6. Sep 6, 2011

### ParoXsitiC

$\sqrt{1+(x/\sqrt{3})^2}$

Shouldnt that be 3 not sqrt(3)

I still don't get it. I need more of a step by step. coming to the solution wolfgram comes to:

I see how you decided what x is, but I dont see how they decided what u was.

7. Sep 6, 2011

### HallsofIvy

No, it shouldn't. Notice the parentheses. Both x and $\sqrt{3}$ are squared.

The crucial point is that you have $\sqrt{1+ a^2}$ and want to get rid of the square root. You should immediately think of the trig identity $1+ tan^2\theta= sec^2\theta$.