How can I integrate x/(sqrrt(1-9x^2)?
Thanks
To integrate $\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx$, you would need to make a u-substitution: $\displaystyle u=1-9x^2$. As a result, $\displaystyle du=-18x\,dx$. Let's substitute all this stuff into the integral now:
$\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx$ now becomes $\displaystyle -\frac{1}{18}\int\frac{\,du}{\sqrt{u}}=-\frac{1}{18}\int u^{-\frac{1}{2}}\,du$.
Integrate:
$\displaystyle -\frac{1}{18}\int u^{-\frac{1}{2}}\,du=-\frac{1}{9}\sqrt{u}+C$.
Substitute u back in and we have:
$\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx=-\frac{1}{9}\sqrt{1-9x^2}+C$
Hope that helped you out!!!