Question

Integrate $ \frac{1}{{{x}^{3}}\,({{x}^{2}}-1)} $

• A. $ \frac{1}{3}\log \left| \frac{{{x}^{3}}}{{{x}^{3}}-1} \right|+C $
• B. $ \frac{1}{3}\log \left| \frac{1-{{x}^{3}}}{{{x}^{3}}} \right|+C $
• C. $ \log \left| \frac{{{x}^{3}}}{{{x}^{3}}-1} \right|+C $
• D. $ \frac{1}{3}\log \left| \frac{{{x}^{3}}-1}{{{x}^{3}}} \right|+C $

Answer 1

Let $ l=\int{\frac{1}{{{x}^{3}}({{x}^{3}}-1)}}\,dx $
$=\int{\frac{1}{{{x}^{3}}-1}}\,dx-\int{\frac{1}{{{x}^{3}}}}\,dx $
$=\int{\frac{1}{(x-1)({{x}^{2}}+x+1)}}dx+\frac{1}{2{{x}^{2}}} $
$=\int{\left[ \frac{1}{3(x-1)}+\frac{\frac{-1}{3}x-\frac{2}{3}}{{{x}^{2}}+x+1} \right]}\,dx+\frac{1}{2{{x}^{2}}} $
$=\int{\frac{1}{3(x-1)}}dx-\frac{1}{3}\int{\frac{x+2}{{{x}^{2}}+x+1}dx+\frac{1}{a{{x}^{2}}}} $
$=\frac{1}{3}\log (x-1)+\frac{1}{2{{x}^{2}}}-\frac{1}{3}\int{\frac{\frac{1}{2}(2x+1)+\frac{3}{2}}{{{x}^{2}}+x+1}}\,dx $
$=\frac{1}{3}\log (x-1)+\frac{1}{2{{x}^{2}}}-\frac{1}{6}\int{\frac{2x+1}{{{x}^{2}}+x+1}}\,dx $ $ -\frac{1}{2}\int{\frac{dx}{{{x}^{2}}+x+1}} $
$=\frac{1}{3}\log (x-1)+\frac{1}{2{{x}^{2}}}-\frac{1}{6}\log \,({{x}^{2}}+x+1) $ $ -\frac{1}{2}\int{\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}} $
$=\frac{1}{3}\log (x-1)+\frac{1}{2{{x}^{2}}}-\frac{1}{6}\log ({{x}^{2}}+x+1) $ $ -\frac{1}{2}.\frac{1}{\frac{\sqrt{3}}{2}}{{\tan }^{-1}}\left( \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)+C $
$=\frac{1}{2}\log (x-1)+\frac{1}{2{{x}^{2}}}-\frac{1}{6}\log ({{x}^{2}}+x+1) $ $ -\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2x+1}{\sqrt{3}} \right)+C $