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Question Number 91326 by mhmd last updated on 29/Apr/20

Commented by mathmax by abdo last updated on 30/Apr/20

$$I=\int \frac{dx}{{(x-1)}^2\sqrt{x^2-x+1}}\Rightarrow I{=}_{x-1=t}\int \frac{dt}{t^2\sqrt{{(t+1)}^2-(t+1)+1}}$$$$=\int \frac{dt}{t^2\sqrt{t^2+2t+1-t-1+1}}=\int \frac{dt}{t^2\sqrt{t^2+t+1}}=\int \frac{dt}{t^2\sqrt{{(t+\frac{1}{2})}^2+\frac{3}{4}}}$$$${=}_{t+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(u\right)}\frac{2}{\sqrt{3}}\int \frac{1}{{(\frac{\sqrt{3}}{2}sh\left(u\right)-\frac{1}{2})}^{2}ch\left(u\right)}\times \frac{\sqrt{3}}{2}ch\left(u\right)du$$$$=4\int \frac{du}{{(\sqrt{3}shu-1)}^2}=4\int \frac{du}{3{sh}^2\left(u\right)-2\sqrt{3}shu+1}$$$$=4\int \frac{du}{\frac{3}{2}(ch\left(2u\right)-1)-2\sqrt{3}shu+1}=8\int \frac{du}{3ch\left(2u\right)-3-4\sqrt{3}shu+2}$$$$=8\int \frac{du}{3sh\left(2u\right)-4\sqrt{3}sh\left(u\right)-1}=8\int \frac{du}{3\frac{e^{2u}-e^{-2u}}{2}-4\sqrt{3}\times \frac{e^u-e^{-u}}{2}-1}$$$$=16\int \frac{du}{3e^{2u}-3e^{-2u}-4\sqrt{3}{e}^u+4\sqrt{3}{e}^{-u}-2}$$$${=}_{e^u=z}16\int \frac{dz}{z(3z^2-3z^{-2}-4\sqrt{3}z+4\sqrt{3}{z}^{-1}-2)}$$$$=16\int \frac{z^{2}dz}{z(3z^4-3-4\sqrt{3}{z}^3+4\sqrt{3}z-2z^2)}$$$$=16\int \frac{z^{2}dz}{z(3z^4-4\sqrt{3}{z}^3-2z^2+4\sqrt{3}z-3)}$$$$decompositionofF\left(z\right)=\frac{z^2}{z(3z^4-4\sqrt{3}{z}^3-2z^2+4\sqrt{3}z-3)}$$$$...becontinued...$$

Answered by MJS last updated on 30/Apr/20

$$2)$$$$r=\sqrt{x^2-x+1}\iff x=\frac{1+\sqrt{4r^2-3}}{2}$$$$r=\frac{\sqrt{3}}{2}\cosh s$$$$s=\ln t$$$$\Rightarrow x=\frac{\sqrt{3}{t}^2+2t-\sqrt{3}}{4t}\iff t=\frac{2x-1+2\sqrt{x^2-x+1}}{\sqrt{3}}$$$$\int \frac{dx}{{(x-1)}^2\sqrt{x^2-x+1}}=$$$$[t=\frac{2x-1+2\sqrt{x^2-x+1}}{\sqrt{3}}\to dx=\frac{\sqrt{3}(t^2+1)}{4t^2}dt]$$$$=\frac{16}{3}\int \frac{t}{{(t-\sqrt{3})}^2{(t+\frac{\sqrt{3}}{3})}^2}dt=$$$$=\sqrt{3}\int \frac{dt}{{(t-\sqrt{3})}^2}-\frac{\sqrt{3}}{3}\int \frac{dt}{{(t+\frac{\sqrt{3}}{3})}^2}-\frac{1}{2}\int \frac{dt}{t-\sqrt{3}}+\frac{1}{2}\int \frac{dt}{t+\frac{\sqrt{3}}{3}}=$$$$=-\frac{\sqrt{3}}{t-\sqrt{3}}+\frac{\sqrt{3}}{3t+\sqrt{3}}-\frac{1}{2}\ln (t-\sqrt{3})+\frac{1}{2}\ln (t+\frac{\sqrt{3}}{3})=$$$$=-\frac{2(\sqrt{3}t+3)}{(t-\sqrt{3})(3t+\sqrt{3})}+\frac{1}{2}\ln \frac{3t+\sqrt{3}}{t-\sqrt{3}}=$$$$=-\frac{\sqrt{x^2-x+1}}{x-1}+\frac{1}{2}\ln \frac{x+1+2\sqrt{x^2-x+1}}{x-1}+C$$

Answered by MJS last updated on 30/Apr/20

$$1)$$$$\underset{0}{\stackrel{\pi }{\int }}\frac{x}{{\csc }^{3}x+{\cot }^{2}x\csc x}dx=$$$$=\underset{0}{\stackrel{\pi }{\int }}\frac{x{\sin }^{3}x}{1+{\cos }^{2}x}dx=$$$$\mathrm{by}\mathrm{parts}$$$$u^{\prime }=\frac{{\sin }^{3}x}{1+{\cos }^{2}x}\to u\stackrel{(\ast )}{=}\cos x-2\arctan \cos x$$$$v=x\to v^{\prime }=1$$$$={\left[x(\cos x-2\arctan \cos x)\right]}_0^{\pi }-\underset{0}{\stackrel{\pi }{\int }}(\cos x-2\arctan \cos x)dx=$$$$[\mathrm{the}\mathrm{integral}\mathrm{is}\mathrm{zero}!]$$$$=\pi (\frac{\pi }{2}-1)$$

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