Question

$\int \frac{1}{(x-1)\sqrt{x^2+x+1}}dx$

• A. $\frac{-1}{\sqrt{3}}\log |(\frac{1}{(x-1)}+\frac{1}{2})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$
• B. $\frac{-1}{2\sqrt{3}}\log |(\frac{1}{(x-1)}+\frac{1}{4})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$
• C. $\frac{-1}{3\sqrt{3}}\log |(\frac{1}{(x-1)}+\frac{1}{2})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$
• D. $\frac{-1}{4\sqrt{3}}\log |(\frac{1}{(x-1)}+\frac{1}{2})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$

Answer 1

The correct option is A $\frac{-1}{\sqrt{3}}\log |(\frac{1}{(x-1)}+\frac{1}{2})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$

We can see that this is another form of irrational algebraic functions. In this we’ll substitute $x-1=\frac{1}{t}$
& $dx=\frac{1}{-t^2}dt$
$\int \frac{1}{(x-1)\sqrt{x^2+x+1}}dx$ will be equal to $=\int \frac{\frac{-1}{t^2}}{\left(\frac{1}{t}\right)\sqrt{{(\frac{1}{t}+1)}^2+(\frac{1}{t}+1)+1}}dt$
$=-\int \frac{1}{\sqrt{3t^2+3t+1}}dt$
$=-\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{(t+\frac{1}{2}{)}^2+\frac{1}{12}}}dt$
We can see that the above for is an standard form of $\int \frac{1}{\sqrt{(x{)}^2+a^2}}dx$
$\int \frac{1}{\sqrt{(x{)}^2+a^2}}dx=\log \left|\right(x+\sqrt{(x{)}^2+a^2}\left)\right|+C$
So, we'll get
$\frac{-1}{\sqrt{3}}\log |(t+\frac{1}{2})+\sqrt{{(t+\frac{1}{2})}^2+\frac{1}{12}}|+C$
On substituting $t=\frac{1}{x-1}$
Or $\frac{-1}{\sqrt{3}}\log |(\frac{1}{x-1}+\frac{1}{2})+\sqrt{\frac{12{(\frac{1}{x-1}+\frac{1}{2})}^2+1}{12}}|+C$