Question
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\frac{1}{\text{t}}$
$\text{dt}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\frac{1}{\text{t}}\sqrt{\Big(\frac{1}{\text{t}^2}+\frac{1}{\text{t}}-1\Big)}}$
$=-\int\frac{\text{dt}}{\sqrt{1+\text{t}-\text{t}^2}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\frac{1}{4}-\text{t}+\text{t}^2\big)}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\text{t}-\frac{1}{2}\big)^2}}$
$=-\sin^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Bigg)+\text{C}$
$\therefore\ \text{I}=-\sin^{-1}\Big(\frac{2\text{t}-1}{\sqrt{5}}\Big)+\text{C}$ $\Big[\text{When}\text{t}=\frac{1}{\text{x}+1}\Big]$
Let $\text{x}+1=\frac{1}{\text{t}}$
$\text{dt}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\frac{1}{\text{t}}\sqrt{\Big(\frac{1}{\text{t}^2}+\frac{1}{\text{t}}-1\Big)}}$
$=-\int\frac{\text{dt}}{\sqrt{1+\text{t}-\text{t}^2}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\frac{1}{4}-\text{t}+\text{t}^2\big)}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\text{t}-\frac{1}{2}\big)^2}}$
$=-\sin^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Bigg)+\text{C}$
$\therefore\ \text{I}=-\sin^{-1}\Big(\frac{2\text{t}-1}{\sqrt{5}}\Big)+\text{C}$ $\Big[\text{When}\text{t}=\frac{1}{\text{x}+1}\Big]$
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