Crystal Wheeler

2022-07-02

Integrate:
$\int \sqrt{{x}^{2}-x+1}dx$

Alexis Fields

$\begin{array}{rl}I& =\int \sqrt{{x}^{2}-x+1}dx\\ & =x\sqrt{{x}^{2}-x+1}-\int x\frac{2x-1}{2\sqrt{{x}^{2}-x+1}}dx\\ & =x\sqrt{{x}^{2}-x+1}-\int \frac{2\left({x}^{2}-x+1\right)+\left(x-\frac{1}{2}\right)-\frac{3}{2}}{2\sqrt{{x}^{2}-x+1}}dx\\ & =x\sqrt{{x}^{2}-x+1}-I-\frac{1}{4}\int \frac{2x-1}{\sqrt{{x}^{2}-x+1}}dx+\frac{3}{4}\int \frac{dx}{\sqrt{{x}^{2}-x+1}}\\ & =\frac{1}{2}\left(x\sqrt{{x}^{2}-x+1}-\frac{1}{2}\sqrt{{x}^{2}-x+1}\right)+\frac{3}{8}\int \frac{dx}{\sqrt{{\left(x-\frac{1}{2}\right)}^{2}+\frac{3}{4}}}\end{array}$
For the last integral, letting $x-\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{tan}\theta$ yields
$\begin{array}{rl}\int \frac{dx}{\sqrt{{\left(x-\frac{1}{2}\right)}^{2}+\frac{3}{4}}}& =\int \frac{\frac{\sqrt{3}}{2}{\mathrm{sec}}^{2}\theta d\theta }{\sqrt{\frac{3}{4}{\mathrm{tan}}^{2}\theta +\frac{3}{4}}}\\ & =\int \mathrm{sec}\theta d\theta \\ & =\mathrm{ln}|\mathrm{sec}\theta +\mathrm{tan}\theta |+{C}_{1}\\ & =\mathrm{ln}|\frac{2\sqrt{{x}^{2}-x+1}}{\sqrt{3}}+\frac{2x-1}{\sqrt{3}}|+{C}_{1}\\ & =\mathrm{ln}|2x-1+2\sqrt{{x}^{2}-x+1}|+{C}_{2}\end{array}$
Now we can conclude that
$I=\frac{2x-1}{4}\sqrt{{x}^{2}-x+1}+\frac{3}{8}\mathrm{ln}|2x-1+2\sqrt{{x}^{2}-x+1}|+C$

Lillianna Andersen

Alternate Method (Using hyperbolic functions)
Letting $x-\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{sinh}\theta$ yields
$\begin{array}{rl}\int \frac{dx}{\sqrt{{\left(x-\frac{1}{2}\right)}^{2}+\frac{3}{4}}}& =\int \frac{\frac{\sqrt{3}}{2}\mathrm{cosh}\theta d\theta }{\sqrt{\frac{3}{4}{\mathrm{sinh}}^{2}\theta +\frac{3}{4}}}\\ & =\int d\theta \\ & =\theta \\ & ={\mathrm{sinh}}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)\end{array}$
$\begin{array}{}& \overline{)I=\frac{2x-1}{4}\sqrt{{x}^{2}-x+1}+\frac{3}{8}{\mathrm{sinh}}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C}\end{array}$

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