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Question Number 125969 by nico last updated on 16/Dec/20

Commented by nico last updated on 16/Dec/20

$$help...$$

Answered by MJS_new last updated on 16/Dec/20

$$\mathrm{let}t=\frac{x}{\sqrt{1-x^2}}\to dx={(1-x^2)}^{3/2}dt$$$$\Rightarrow$$$$\int \frac{t+1}{(t^2+1)(t^2-t+1)}dt=\int \frac{t-1}{t^2+1}dt-\int \frac{t-2}{t^2-t+1}dt=$$$$=\frac{1}{2}\int \frac{2t}{t^2+1}dt-\int \frac{dt}{t^2+1}-\frac{1}{2}\int \frac{2t-1}{t^2-t+1}dt+6\int \frac{dt}{{(2t-1)}^2+3}=$$$$=\frac{1}{2}\ln (t^2+1)-\arctan t-\frac{1}{2}\ln (t^2-t+1)+\sqrt{3}\arctan \frac{\sqrt{3}(2t-1)}{3}=$$$$...$$$$=\sqrt{3}\arctan \frac{\sqrt{3}(2x-\sqrt{1-x^2})}{3\sqrt{1-x^2}}-\arcsin x-\frac{1}{2}\ln \mid 1-x\sqrt{1-x^2}\mid +C$$

Answered by mathmax by abdo last updated on 16/Dec/20

$$\mathrm{A}=\int \frac{\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}}{1-\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{ch}.\sqrt{1-{\mathrm{x}}^2}=1-\mathrm{xt}\Rightarrow 1-{\mathrm{x}}^2={(1-\mathrm{xt})}^2$$$$\Rightarrow 1-{\mathrm{x}}^2=1-2\mathrm{xt}+{\mathrm{x}}^2{\mathrm{t}}^2\Rightarrow -{\mathrm{x}}^2=-2\mathrm{xt}+{\mathrm{x}}^2{\mathrm{t}}^2\Rightarrow -\mathrm{x}=-2\mathrm{t}+{\mathrm{xt}}^2$$$$\Rightarrow -2\mathrm{t}+{\mathrm{xt}}^2+\mathrm{x}=0\Rightarrow (1+{\mathrm{t}}^2)\mathrm{x}=2\mathrm{t}\Rightarrow \mathrm{x}=\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}\Rightarrow$$$$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2({\mathrm{t}}^2+1)-2\mathrm{t}\left(2\mathrm{t}\right)}{{({\mathrm{t}}^2+1)}^2}=\frac{{2\mathrm{t}}^2+2-{4\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^2}=\frac{2-{2\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^2}\Rightarrow$$$$\mathrm{A}=\int \frac{\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}+1-\frac{{2\mathrm{t}}^2}{{\mathrm{t}}^2+1}}{1-\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}(1-\frac{{2\mathrm{t}}^2}{{\mathrm{t}}^2+1})}\times \frac{2-{2\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt}$$$$=\int \frac{2\mathrm{t}+{\mathrm{t}}^2+1-{2\mathrm{t}}^2}{{({\mathrm{t}}^2+1)}^3(1-\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}\times \frac{1-{\mathrm{t}}^2}{{\mathrm{t}}^2+1})}\mathrm{dt}$$$$=\int \frac{-{\mathrm{t}}^2+2\mathrm{t}+1}{{({\mathrm{t}}^2+1)}^3(1-\frac{2\mathrm{t}-{2\mathrm{t}}^3}{{({\mathrm{t}}^2+1)}^2})}\mathrm{dt}$$$$=\int \frac{-{\mathrm{t}}^2+2\mathrm{t}+1}{({\mathrm{t}}^2+1)({\mathrm{t}}^4+{2\mathrm{t}}^2+1-2\mathrm{t}+{2\mathrm{t}}^3)}\mathrm{dt}$$$$=\int \frac{-{\mathrm{t}}^2+2\mathrm{t}+1}{({\mathrm{t}}^2+1)({\mathrm{t}}^4+{2\mathrm{t}}^3+{2\mathrm{t}}^2-2\mathrm{t}+1)}\mathrm{dt}\mathrm{rest}\mathrm{decomposition}\mathrm{of}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{-{\mathrm{t}}^2+2\mathrm{t}+1}{({\mathrm{t}}^2+1)({\mathrm{t}}^4+{2\mathrm{t}}^3+{2\mathrm{t}}^2-2\mathrm{t}+1)}....\mathrm{be}\mathrm{continued}...$$

Answered by liberty last updated on 17/Dec/20

$$let{x}^2=\frac{p^2}{1+p^2}\wedge \frac{1}{x^2}=\frac{1+p^2}{p^2}\Rightarrow 2xdx=\frac{2p(1+p^2)-2p^3}{{(1+p^2)}^2}dp$$$$xdx=\frac{p}{{(1+p^2)}^2}dp\Rightarrow dx=\frac{dp}{{(1+p^2)}^{3/2}}$$$$I=\int \frac{\frac{p}{\sqrt{1+p^2}}+1}{1-\frac{p}{\sqrt{1+p^2}}}\left(\frac{dp}{{(1+p^2)}^{3/2}}\right)$$$$I=\int \frac{p+\sqrt{1+p^2}}{\sqrt{1+p^2}-p}\left(\frac{dp}{{(1+p^2)}^{3/2}}\right)$$$$letp=\tan z$$$$I=\int \left(\frac{\tan z+\sec z}{\sec z-\tan z}\right)\left(\frac{\sec {}^{2}z}{\sec {}^{3}z}\right)dz$$$$I=\int \left(\frac{\sin z+1}{1-\sin z}\right)\cos zdz$$$$I=\int \left(\frac{1+2\sin z+\sin {}^{2}z}{\cos z}\right)dz$$$$I=\int \left(\frac{2+2\sin z-\cos {}^{2}z}{\cos z}\right)dz=\int (2\sec z-\cos z+\frac{2\sin z}{\cos z})dz$$$$=2\ln \mid \sec z+\tan z\mid -\sin z-2\ln \mid \cos z\mid +c$$$$=2\ln \mid \frac{\sec z+\tan z}{\cos z}\mid -\sin z+c$$$$=2\ln \mid \frac{1+\sin z}{\cos {}^{2}z}\mid -\sin z+c$$$$=2\ln \mid \frac{1+\frac{p}{\sqrt{1+p^2}}}{\frac{1}{1+p^2}}\mid -\frac{p}{\sqrt{1+p^2}}+c$$$$=2\ln \mid 1+p^2+p\sqrt{1+p^2}\mid -\frac{p}{\sqrt{1+p^2}}+c$$$$=-x+2\ln \mid 1+\frac{x^2}{1-x^2}+\frac{x}{1-x^2}\mid +c$$$$=-x+2\ln \mid \frac{1}{1-x}\mid +c$$

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