Question
$\int \frac{ d x}{x^3-1}$
✓
Answer
$ \text { Let } I =\int \frac{ d x}{x^3-1}$
$=\int \frac{1}{(x-1)\left(x^2+x+1\right)} d x $
$ \text { Let } \frac{1}{(x-1)\left(x^2+x+1\right)}$
$=\frac{ A }{x-1}+\frac{ B x+ C }{x^2+x+1}$
$\therefore 1= A \left( x ^2+ x +1\right)+( Bx + C )( x -1)\ldots(i) $
Putting $x=1$ in (i), we get
$ 1= A \left(1^2+1+1\right)$
$\therefore 1=3 A$
$\therefore A =\frac{1}{3} $
Putting $x=0$ in (i), we get
$ 1= A (0+0+1)+(0+ C )(0-1)$
$\therefore 1= A - C$
$\therefore 1=\frac{1}{3}- C$
$\therefore C =-\frac{2}{3} $
Putting $x=2$ in (i), we get
$ 1= A \left(2^2+2+1\right)+(2 B + C )(2-1)$
$\therefore 1=7 A +2 B + C$
$\therefore 1=\frac{7}{3}+2 B -\frac{2}{3}$
$\therefore 1=\frac{5}{3}+2 B$
$\therefore \frac{-2}{3}=2 B$
$\therefore B =-\frac{1}{3}$
$\therefore I=\int\left(\frac{\frac{1}{3}}{x-1}+\frac{-\frac{1}{3} x-\frac{2}{3}}{x^2+x+1}\right) d x$
$=\frac{1}{3} \int\left(\frac{1}{x-1}-\frac{x+2}{x^2+x+1}\right) d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x+2}{x^2+x+1} d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \cdot \frac{1}{2} \int \frac{2 x+4}{x^2+x+1} d x$
$=\frac{1}{3} \int \frac{1}{x 1} d x-\frac{1}{6} \int \frac{(2 x+1)+3}{x^2+x+1} \cdot d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{6} \int \frac{2 x+1}{x^2+x+1} d x-\frac{1}{2} \int \frac{ d x}{x^2+x+1}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \int \frac{ d x}{x^2+x+\frac{1}{4}-\frac{1}{4}+1}$
${\left[\because \int \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right]}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \int \frac{ d x}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+ c$
$\therefore I =\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+ c $
$=\int \frac{1}{(x-1)\left(x^2+x+1\right)} d x $
$ \text { Let } \frac{1}{(x-1)\left(x^2+x+1\right)}$
$=\frac{ A }{x-1}+\frac{ B x+ C }{x^2+x+1}$
$\therefore 1= A \left( x ^2+ x +1\right)+( Bx + C )( x -1)\ldots(i) $
Putting $x=1$ in (i), we get
$ 1= A \left(1^2+1+1\right)$
$\therefore 1=3 A$
$\therefore A =\frac{1}{3} $
Putting $x=0$ in (i), we get
$ 1= A (0+0+1)+(0+ C )(0-1)$
$\therefore 1= A - C$
$\therefore 1=\frac{1}{3}- C$
$\therefore C =-\frac{2}{3} $
Putting $x=2$ in (i), we get
$ 1= A \left(2^2+2+1\right)+(2 B + C )(2-1)$
$\therefore 1=7 A +2 B + C$
$\therefore 1=\frac{7}{3}+2 B -\frac{2}{3}$
$\therefore 1=\frac{5}{3}+2 B$
$\therefore \frac{-2}{3}=2 B$
$\therefore B =-\frac{1}{3}$
$\therefore I=\int\left(\frac{\frac{1}{3}}{x-1}+\frac{-\frac{1}{3} x-\frac{2}{3}}{x^2+x+1}\right) d x$
$=\frac{1}{3} \int\left(\frac{1}{x-1}-\frac{x+2}{x^2+x+1}\right) d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x+2}{x^2+x+1} d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \cdot \frac{1}{2} \int \frac{2 x+4}{x^2+x+1} d x$
$=\frac{1}{3} \int \frac{1}{x 1} d x-\frac{1}{6} \int \frac{(2 x+1)+3}{x^2+x+1} \cdot d x$
$=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{6} \int \frac{2 x+1}{x^2+x+1} d x-\frac{1}{2} \int \frac{ d x}{x^2+x+1}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \int \frac{ d x}{x^2+x+\frac{1}{4}-\frac{1}{4}+1}$
${\left[\because \int \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right]}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \int \frac{ d x}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+ c$
$\therefore I =\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^2+x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+ c $
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