Integral Evaluation: X^2/((x-1/3)^2-16/9)^(3/2)

[danield]

Homework Statement

Evaluate the integral.
http://www.webassign.net/www30/symImages/7/a/8fcca85e1856587219a0f28aa29d3a.gif

Homework Equations

I think that i need to complete the square and then use, x=sec u, dx=sec u tan u du

The Attempt at a Solution

i have attempted various methods but none has yield the right result :S
ive gotten
X^2/((x-1/3)^2-16/9)^(3/2)

then i used U=x-1/3
(u+1/3)^2/(U^2-16/9)^(3/2)
then i use u=(4/3)sec o and du=(4/3)sec o tan o do

and i get

((16/9)(sec o )^2 + (8/9)sec o + (1/9))/(64/27)(Tan o)^3

and i don't know where to go from there b/c the methods i have used have not worked

 

[Nino MMP]

A:We have $$\int \frac{x^2}{(x-\frac{1}{3})^{\frac{3}{2}}\sqrt{\frac{16}{9}+\left(x-\frac{1}{3}\right)^2}}dx$$Let $x-\frac{1}{3}=\frac{4}{3}\sec\theta$, then $\frac{dx}{d\theta}=\frac{4}{3}\sec\theta\tan\theta$.So we get $$\frac{4}{3}\int\frac{\sec^2\theta}{\tan^3\theta\sqrt{\sec^2\theta+\frac{16}{9}}}d\theta$$Let $u=\tan\theta$, then $du=\sec^2\theta d\theta$. We get $$\frac{4}{3}\int\frac{1}{u^3\sqrt{u^2+\frac{16}{9}}}du$$Integrate it, and you can get the answer.