# Definite Integral

1. May 26, 2005

### cogs24

hi guys
yeh, im still going through revision, and im also stuck on this question.

*integral sign*(upper limit 4, lower limit 2) x/sqrt(3x^2 + 4).dx

When i look at this, i think of letting u = 3x^2 + 4, then bringing that sqrt to the top,solving for dx, and then find the integral, and sub the values in to find the definite answer
Could someone clarify this for me
Thanx

2. May 26, 2005

### whozum

a U substitution would take care of it, but you would still be left with [itex] u^{-\frac{1}{2}}[/tex], so you wouldnt be bringing the sqrt to the top unless you rationalized the denominator.

$$\int \frac{x}{\sqrt{3x^2+4}}\ dx = \frac{1}{6} \int \frac{1}{\sqrt{u}} \ du \ where \ u = 3x^2+4$$

3. May 26, 2005

### James R

Your approach is basically right. You want

$$\int^4_2\frac{ x}{\sqrt{3x^2 + 4}} dx$$

Put

$$u= 3x^2 + 4$$

and sub it in. Don't forget about changing the dx to a du in the integral.