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Question Number 168745 by MikeH last updated on 17/Apr/22

$$\int \frac{1}{x+\sqrt{x-1}}dx=??$$

Commented by safojontoshtemirov last updated on 17/Apr/22

$$\int \frac{dx}{x+\sqrt{x-1}}=\sqrt{x-1}=t;x=t^2+1;dx=2tdt$$$$\int \frac{2tdt}{t^2+t+1}=\int \frac{2t+1}{t^2+t+1}dt-\int \frac{1}{t^2+t+1}dt=I_1-I_2$$$$I_1=\int \frac{d(t^2+t+1)}{t^2+t+1}dt=ln(t^2+t+1)+C_1=ln(x-1+\sqrt{x-1}+1)+C_1$$$$I_2=\int \frac{1}{t^2+t+1}dt=\int \frac{1}{{(t+\frac{1}{2})}^2+\frac{3}{4}}dt=\frac{4}{3}\int \frac{1}{{\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)}^2+1}dt=$$$$\frac{4}{3}\int \frac{1}{{\left(\frac{2t+1}{\sqrt{3}}\right)}^2+1}dt=\frac{4}{3}\cdot \frac{\sqrt{3}}{2}\int \frac{d\left(\frac{2t+1}{\sqrt{3}}\right)}{{\left(\frac{2t+1}{\sqrt{3}}\right)}^2+1}=$$$$\frac{1}{\sqrt{3}}arctg\left(\frac{2t+1}{\sqrt{3}}\right)+C_2=\frac{2}{\sqrt{3}}arctg\frac{2\sqrt{x-1}+1}{\sqrt{3}}+C_2$$$$I_1-I_2=ln(x-1+\sqrt{x-1}+1)-\frac{2}{\sqrt{3}}arctg\frac{2\sqrt{x-1}+1}{\sqrt{3}}+C$$$$$$

Answered by blackmamba last updated on 17/Apr/22

$$=\int \frac{x-\sqrt{x-1}}{x^2-x+1}dx$$$$=\int \frac{\frac{1}{2}(2x-1)+\frac{1}{2}}{x^2-x+1}dx-\int \frac{\sqrt{x-1}}{x^2-x+1}dx$$$$=\frac{1}{2}\ln (x^2-x+1)+\frac{1}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}-\int \frac{\sqrt{x-1}}{x^2-x+1}dx$$

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