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Question Number 107567 by Ar Brandon last updated on 11/Aug/20

$$\int \sqrt{{3\mathrm{x}}^2-2\mathrm{x}}\mathrm{dx}$$

Answered by bobhans last updated on 11/Aug/20

$$\int \sqrt{{3\mathrm{x}}^2-2\mathrm{x}}\mathrm{dx}=\int \sqrt{3({\mathrm{x}}^2-\frac{2\mathrm{x}}{3}+\frac{1}{9})-\frac{1}{3}}\mathrm{dx}$$$$=\sqrt{3}\int \sqrt{{(\mathrm{x}-\frac{1}{3})}^2-{\left(\frac{1}{\sqrt{3}}\right)}^2}\mathrm{dx}$$$$\mathrm{set}\mathrm{x}-\frac{1}{3}=\frac{1}{\sqrt{3}}\sec \theta$$$$=\sqrt{3}\int \sqrt{\frac{1}{3}\tan {}^2\theta }\frac{1}{\sqrt{3}}\sec \theta \tan \theta \mathrm{d}\theta$$$$=\frac{\sqrt{3}}{3}\int \tan {}^2\theta \sec \theta \mathrm{d}\theta =\frac{\sqrt{3}}{3}\int (\sec {}^3\theta -\sec \theta )\mathrm{d}\theta$$$$\mathrm{note}\int \sec {}^3\mathrm{x}\mathrm{dx}=\frac{\sin \mathrm{x}}{2\cos {}^2\mathrm{x}}+\frac{1}{2}\ln \mid \tan (\mathrm{x}+\frac{\pi }{4})\mid +\mathrm{c}$$$$=\frac{1}{2}\tan \mathrm{x}\sec \mathrm{x}+\frac{1}{2}\ln \mid \frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}\mid +\mathrm{c}$$$$\mathrm{and}\int \sec \mathrm{x}\mathrm{dx}=\ln \mid \sec \mathrm{x}+\tan \mathrm{x}\mid +\mathrm{c}$$$$$$

Commented by Ar Brandon last updated on 11/Aug/20

Thanks Sir.��

Commented by bobhans last updated on 12/Aug/20

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Answered by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{I}=\int \sqrt{3({\mathrm{x}}^2-\frac{2}{3}\mathrm{x})}\mathrm{dx}=\sqrt{3}\int \sqrt{{\mathrm{x}}^2-\frac{2}{3}\mathrm{x}+\frac{1}{9}-\frac{1}{9}}\mathrm{dx}$$$$=\sqrt{3}\int \sqrt{{(\mathrm{x}-\frac{1}{3})}^2-\frac{1}{9}}\mathrm{dx}{=}_{\mathrm{x}-\frac{1}{3}=\frac{1}{3}\mathrm{ch}\left(\mathrm{t}\right)}\sqrt{3}\int \frac{1}{3}\mathrm{sht}.\frac{1}{3}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{dt}$$$$=\frac{\sqrt{3}}{9}\int {\mathrm{sh}}^2\mathrm{t}\mathrm{dt}=\frac{\sqrt{3}}{9}\int \frac{\mathrm{ch}\left(2\mathrm{t}\right)-1}{2}\mathrm{dt}=\frac{\sqrt{3}}{18}\int \mathrm{ch}\left(2\mathrm{t}\right)\mathrm{dt}-\frac{\sqrt{3}}{18}\mathrm{t}$$$$=\frac{\sqrt{3}}{36}\mathrm{sh}\left(2\mathrm{t}\right)-\frac{\sqrt{3}}{18}\mathrm{t}+\mathrm{C}$$$$=\frac{\sqrt{3}}{18}\mathrm{ch}\left(\mathrm{t}\right)\mathrm{sh}\left(\mathrm{t}\right)-\frac{\sqrt{3}}{18}\mathrm{t}+\mathrm{C}$$$$=\frac{\sqrt{3}}{18}(3\mathrm{x}-1)\sqrt{{(3\mathrm{x}-1)}^2-1}-\frac{\sqrt{3}}{18}\mathrm{argch}(3\mathrm{x}-1)+\mathrm{C}$$$$\mathrm{I}=\frac{\sqrt{3}}{18}(3\mathrm{x}-1)\sqrt{{(3\mathrm{x}-1)}^2-1}-\frac{\sqrt{3}}{18}\ln (3\mathrm{x}-1+\sqrt{{(3\mathrm{x}-1)}^2-1})+\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 12/Aug/20

$$\sqrt{3}\int \sqrt{x^2-\frac{2}{3}x}dx$$$$\sqrt{3}\int \sqrt{{(x-\frac{1}{3})}^2-{\left(\frac{1}{3}\right)}^2}dx$$$$\sqrt{3}\frac{x-\frac{1}{3}}{2}\sqrt{x^2-\frac{2}{3}x}-\sqrt{3}.\frac{1}{18}log(x-\frac{1}{3}+\sqrt{x^2-\frac{2}{3}x})$$$$\frac{\sqrt{3}}{18}(\frac{3x-1}{\sqrt{3}}\sqrt{3x^2-2x}-log(x-\frac{1}{3}+\sqrt{x^2-\frac{2}{3}x}))+C$$

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