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Question Number 122397 by Ar Brandon last updated on 16/Nov/20

$$\mathrm{a}\mathrm{\setminus }\int \frac{2\mathrm{x}+1}{1+\mathrm{x}}\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\mathrm{dx}\mathrm{c}\mathrm{\setminus }\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}+\sqrt[3]{\mathrm{x}}}$$$$\mathrm{b}\mathrm{\setminus }\int \frac{\mathrm{dx}}{\mathrm{x}+\sqrt{\mathrm{x}-1}}\mathrm{d}\mathrm{\setminus }\int \frac{\mathrm{dx}}{(1+\mathrm{x})\sqrt{1+\mathrm{x}+{\mathrm{x}}^2}}$$

Answered by Dwaipayan Shikari last updated on 16/Nov/20

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

Answered by Dwaipayan Shikari last updated on 16/Nov/20

$$\int \frac{dx}{\sqrt[3]{x}(1+\sqrt[6]{x})}x=t^6\Rightarrow 1=6t^5\frac{dt}{dx}$$$$=\int \frac{6t^5}{t^2(1+t)}dt=6\int t^2-t+1-\frac{1}{t+1}dt=2t^3-3t^2+6t-6log(1+t)+C$$$$=2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6log(1+\sqrt[6]{x})+C$$

Answered by Dwaipayan Shikari last updated on 16/Nov/20

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

Commented by Ar Brandon last updated on 16/Nov/20

Nice path. Thanks

Answered by MJS_new last updated on 17/Nov/20

$$\int \frac{2x+1}{1+x}\sqrt{\frac{1-x}{1+x}}dx=$$$$[t=\sqrt{\frac{1-x}{1+x}}\to dx=-\sqrt{1-x}{(1+x)}^{3/2}dt]$$$$=2\int \frac{t^2(t^2-3)}{{(t^2+1)}^2}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{2t(t^2+3)}{t^2+1}-6\int \frac{dt}{t^2+1}=$$$$=\frac{2t(t^2+3)}{t^2+1}-6\arctan t=$$$$=2(x+2)\sqrt{\frac{1-x}{1+x}}-6\arctan \sqrt{\frac{1-x}{1+x}}+C$$$$$$$$\int \frac{dx}{x+\sqrt{x-1}}=$$$$[t=\sqrt{x-1}\to dx=2tdt]$$$$=2\int \frac{t}{t^2+t+1}dt=\int (\frac{2t+1}{t^2+t+1}-\frac{1}{t^2+t+1})dt=$$$$=\ln (t^2+t+1)-\frac{2\sqrt{3}}{3}\arctan \frac{(2t+1)\sqrt{3}}{3}=$$$$=\ln (x+\sqrt{x-1})-\frac{2\sqrt{3}}{3}\arctan \frac{(1+2\sqrt{x-1})\sqrt{3}}{3}+C$$$$$$$$\int \frac{dx}{\sqrt{x}+\sqrt[3]{x}}=\int \frac{dx}{x^{1/3}(x^{1/6}+1)}=$$$$[t=x^{1/6}\to dx=6x^{5/6}dt]$$$$=6\int \frac{t^3}{t+1}dt=6\int (t^2-t+1-\frac{1}{t+1})dt=$$$$=2t^3-3t^2+6t-6\ln (t+1)=$$$$=2x^{1/2}-3x^{1/3}+6x^{1/6}-6\ln (x^{1/6}+1)+C$$$$$$$$\int \frac{dx}{(1+x)\sqrt{1+x+x^2}}=$$$$[t=\frac{2x+1+2\sqrt{x^2+x+1}}{\sqrt{3}}\to dx=\frac{\sqrt{3(x^2+x+1)}}{2x+1+\sqrt{x^2+x+1}}dt]$$$$=4\sqrt{3}\int \frac{dt}{3t^2+2\sqrt{3}t-3}=$$$$=\int (\frac{1}{t-\frac{\sqrt{3}}{3}}-\frac{1}{t+\sqrt{3}})dt=$$$$=\ln \frac{t-\frac{\sqrt{3}}{3}}{t+\sqrt{3}}=\ln \frac{x-1+2\sqrt{x^2+x+1}}{x+1}+C$$

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