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Question Number 46020 by mondodotto@gmail.com last updated on 20/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

$$=\frac{1}{2}\int \frac{x^2+x+1}{x\sqrt{x^2+x+1}}dx$$$$=\frac{1}{2}\int \frac{x}{\sqrt{x^2+x+1}}+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}+\frac{1}{2}\int \frac{dx}{x\sqrt{x^2+x+1}}$$$$=\frac{1}{4}\int \frac{2x+1-1}{\sqrt{x^2+x+1}}+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}+\frac{1}{2}\int \frac{dx}{x\sqrt{x^2+x+1}}$$$$=\frac{1}{4}\int \frac{d(x^2+x+1)}{\sqrt{x^2+x+1}}+(\frac{1}{2}-\frac{1}{4})\int \frac{dx}{\sqrt{x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}}+\frac{1}{2}\int \frac{dx}{x\sqrt{x^2+x+1}}$$$$I_1+I_2+I_3$$$$I_1=\frac{1}{4}\times \frac{{(x^2+x+1)}^{\frac{-1}{2}+1}}{\frac{1}{2}}+c_1$$$$I_1=\frac{1}{2}\times \sqrt{x^2+x+1}+c_1\leftarrow I_1$$$$I_2=\frac{1}{4}\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$I_2=\frac{1}{4}ln\{(x+\frac{1}{2})+\sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\}+c_2$$$$=\frac{1}{4}ln\{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+c_2$$$$I_3=\frac{1}{2}\int \frac{dx}{x\sqrt{x^2+x+1}}$$$$letx=\frac{1}{t}dx=-\frac{1}{t^2}dt$$$$=\frac{1}{2}\int \frac{-dt}{t^2\times \frac{1}{t}\times \sqrt{\frac{1}{t^2}+\frac{1}{t}+1}}$$$$=\frac{-1}{2}\int \frac{tdt}{t\sqrt{1+t+t^2}}$$$$=\frac{-1}{2}\int \frac{dt}{\sqrt{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$\frac{-1}{2}ln\{(t+\frac{1}{2})+\sqrt{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\}+c_3$$$$=\frac{-1}{2}ln\{(\frac{1}{x}+\frac{1}{2})+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\}+c_3$$

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