1. $\int \frac{1}{x\sqrt{x^3-1}}dx$
Note that, $I = \int \frac{x^2}{x^3\sqrt{x^3-1}}dx$
Hence $du=3x^2 \ dx$.
As a result we have,
\begin{eqnarray}
I&=&1/3\int\frac{du}{\sqrt{u}(u+1)}\\
&=& 2/3\int \frac{v\ dv}{v(v^2+1)} \mbox { (Substituting } u=v^2 \mbox{we get } du = 2v\ dv)\\
&=&(2/3)\tan^{-1}(\sqrt{x^3-1})
\end{eqnarray}
No comments:
Post a Comment