Solving Definite Integral: x/sqrt(3x^2 + 4)

[cogs24]

hi guys
yeh, I am still going through revision, and I am 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 into find the definite answer
Could someone clarify this for me
Thanx

 

[whozum]

a U substitution would take care of it, but you would still be left with [itex]u^{-\frac{1}{2}}[/tex], so you wouldn't be bringing the sqrt to the top unless you rationalized the denominator. <br /> <br /> $$\int \frac{x}{\sqrt{3x^2+4}}\ dx = \frac{1}{6} \int \frac{1}{\sqrt{u}} \ du \ where \ u = 3x^2+4$$<br /> <br /> Dont forget to adjust your integration limits.[/itex]

 

[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.