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Question Number 93820 by i jagooll last updated on 15/May/20

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

Commented by mathmax by abdo last updated on 17/May/20

$$I=\int \frac{x^2}{\sqrt{x^2-x+1}}dx\Rightarrow I=\int \frac{x^2-x+1}{\sqrt{x^2-x+1}}dx+\int \frac{x-1}{\sqrt{x^2-x+1}}dx=I_1+I_2$$$$\int \frac{x^2-x+1}{\sqrt{x^2-x+1}}dx=\int \sqrt{x^2-x+1}dx=\int \sqrt{{(x-\frac{1}{2})}^2+\frac{3}{4}}dx$$$${=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)}\frac{\sqrt{3}}{2}\int ch\left(t\right)\times \frac{\sqrt{3}}{2}ch\left(t\right)dt=\frac{3}{4}\int \left(\frac{1+ch\left(2t\right)}{2}\right)dt$$$$=\frac{3}{8}t+\frac{3}{16}sh\left(2t\right)+c=\frac{3}{8}t+\frac{3}{8}sh\left(t\right)ch\left(t\right)+c$$$$=\frac{3}{8}argsh\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{3}{8}\left(\frac{2x-1}{3}\right)\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2}$$$$I_2=\frac{1}{2}\int \frac{2x-1-1}{\sqrt{x^2-x+1}}dx=\sqrt{x^2-x+1}-\frac{1}{2}\int \frac{dx}{\sqrt{x^2-x+1}}$$$$\int \frac{dx}{\sqrt{x^2-x+1}}{=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}t}\frac{\sqrt{3}}{2}\int \frac{dt}{\frac{\sqrt{3}}{2}\sqrt{1+t^2}}=\int \frac{dt}{\sqrt{1+t^2}}=ln(t+\sqrt{1+t^2})$$$$=ln(\frac{2x-1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2})\Rightarrow$$$$I=\frac{3}{8}ln(\frac{2x-1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2})+\frac{2x-1}{8}\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2}$$$$+\sqrt{x^2-x+1}-\frac{1}{2}ln(\frac{2x-1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2})+C$$$$=-\frac{1}{8}ln(\frac{2x-1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2})+\sqrt{x^2-x+1}+\frac{2x-1}{8}\sqrt{1+{\left(\frac{2x-1}{\sqrt{3}}\right)}^2}+C$$

Answered by Kunal12588 last updated on 15/May/20

$$I=\int \frac{x^{2}dx}{\sqrt{x^2-x+1}}$$$$=\int \frac{x^2-x+1}{\sqrt{x^2-x+1}}dx+\int \frac{x-1}{\sqrt{x^2-x+1}}dx$$$$=\int \sqrt{x^2-x+1}dx+\frac{1}{2}\int \frac{2x-1}{\sqrt{x^2-x+1}}dx-\frac{1}{2}\int \frac{dx}{\sqrt{x^2-x+1}}$$$$=\int \sqrt{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx+\frac{1}{2}\int \frac{d(x^2-x+1)}{\sqrt{x^2-x+1}}$$$$-\frac{1}{2}\int \frac{dx}{\sqrt{(x-\frac{1}{2})+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$=\frac{1}{2}(x-\frac{1}{2})\sqrt{x^2-x+1}+\frac{3}{8}ln\mid x-\frac{1}{2}+\sqrt{x^2-x+1}\mid$$$$+\frac{1}{2}\times 2\sqrt{x^2-x+1}-\frac{1}{2}\times \frac{2}{\sqrt{3}}{tan}^{-1}\frac{2x-1}{\sqrt{3}}+C$$$$=\frac{1}{4}(2x-1)\sqrt{x^2-x+1}+\frac{3}{8}ln\mid x-\frac{1}{2}+\sqrt{x^2-x+1}\mid$$$$+\sqrt{x^2-x+1}-\frac{1}{\sqrt{3}}{tan}^{-1}\frac{2x-1}{\sqrt{3}}+C$$$$=\frac{1}{4}(2x+3)\sqrt{x^2-x+1}+\frac{3}{8}ln\mid x-\frac{1}{2}+\sqrt{x^2-x+1}\mid$$$$-\frac{1}{\sqrt{3}}{tan}^{-1}\frac{2x-1}{\sqrt{3}}+C$$

Commented by i jagooll last updated on 15/May/20

$$\mathrm{waw}..\mathrm{great}\mathrm{sir}$$

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