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Question Number 167330 by LEKOUMA last updated on 13/Mar/22

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

Answered by greogoury55 last updated on 13/Mar/22

$$=\int \frac{\sqrt{x^2+x+1}-x}{x+1}dx$$$$=\int \frac{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}-x}{x+1}dx$$$$K_1=\int \frac{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}{x+1}dx$$$$[x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan t]$$$$K_1=\int \frac{\frac{\sqrt{3}}{2}\sec t}{\frac{1}{2}(1+\sqrt{3}\tan t)}\frac{\sqrt{3}}{2}.{\sec }^{2}tdt$$$$K_1=\frac{3}{2}\int \frac{\sec {}^{2}t}{\cos t+\sqrt{3}\sin t}dt$$$$K_1=\frac{3}{2}\int \frac{\sec {}^{2}t}{2(\frac{1}{2}\cos t+\frac{\sqrt{3}}{2}\sin t)}dt$$$$K_1=\frac{3}{4}\int \frac{\sec {}^{2}t}{\cos (t-\frac{\pi }{3})}dt$$$$K_2=\int \frac{x}{x+1}dx=\int \frac{(x+1)-1}{x+1}dx$$$$K_2=x-\ln \mid x+1\mid +c$$

Commented by peter frank last updated on 14/Mar/22

$$\mathrm{thank}\mathrm{you}$$

Answered by MJS_new last updated on 13/Mar/22

$$\int \frac{dx}{x+\sqrt{x^2+x+1}}=-\int \frac{x-\sqrt{x^2+x+1}}{x+1}dx=$$$$[t=x+\frac{1}{2}\to dx=dt]$$$$=-\int \frac{2t-1-\sqrt{4t^2+3}}{2t+1}dt=$$$$[u=\frac{\sqrt{3}}{3}(2t+\sqrt{4t^2+3})\to dt=\frac{\sqrt{3}(u^2+1)}{4u^2}]$$$$=\frac{\sqrt{3}}{2}\int \frac{u^2+1}{u^2(\sqrt{3}u-1)}du=$$$$=\int (\frac{2\sqrt{3}}{\sqrt{3}u-1}-\frac{3}{2u}-\frac{\sqrt{3}}{2u^2})du=$$$$=2\ln (\sqrt{3}u-1)-\frac{3}{2}\ln u+\frac{\sqrt{3}}{2u}=$$$$...$$$$=2\ln (x+\sqrt{x^2+x+1})-\frac{3}{2}\ln (2x+1+\sqrt{x^2+x+1})-x+\sqrt{x^2+x+1}+C$$

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