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Question Number 30182 by abdo imad last updated on 17/Feb/18

$$find{\int }_2^3\frac{\sqrt{x+1}}{x\sqrt{1-x}}dx.$$

Commented by abdo imad last updated on 21/Feb/18

$$letput\sqrt{\frac{x+1}{x-1}}=t\iff \frac{x+1}{x-1}=t^2\iff x+1=-t^2+t^{2}x\iff$$$$(1-t^2)x=-t^2-1\iff x=\frac{-t^2-1}{-t^2+1}=\frac{t^2+1}{t^2-1}=1+\frac{2}{t^2-1}\Rightarrow$$$$\frac{dx}{dt}=\frac{-4t}{{(t^2-1)}^2}\Rightarrow I={\int }_{\sqrt{3}}^{\sqrt{2}}t\frac{t^2-1}{t^2+1}\frac{-4t}{{(t^2-1)}^2}dt$$$$=4{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{t^2}{(t^2+1)(t^2-1)}dt=4{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{t^2-1+1}{(t^2+1)(t^2-1)}dt$$$$=4{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{t^2+1}+4{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{(t^2+1)(t^2-1)}wehave$$$${\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{1+t^2}=arctan\left(\sqrt{3}\right)-arctan\left(\sqrt{2}\right)and$$$${\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{(t^2+1)(t^2-1)}=\frac{1}{2}{\int }_{\sqrt{2}}^{\sqrt{3}}(\frac{1}{t^2-1}-\frac{1}{t^2+1})dt$$$$=\frac{1}{2}{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{t^2-1}-\frac{1}{2}\left(arctan(\sqrt{3)}-arctan\sqrt{2})\right)$$$$=\frac{1}{2}{\int }_{\sqrt{2}}^{\sqrt{3}}(\frac{1}{t-1}-\frac{1}{t+1})dt-\frac{1}{2}(....)$$$$=\frac{1}{2}{[ln\mid \frac{t-1}{t+1}\mid ]}_{\sqrt{2}}^{\sqrt{3}}-\frac{1}{2}(...)\Rightarrow$$$$I=4(artan\left(\sqrt{3}\right)-arctan\left(\sqrt{2}\right))+2{\int }_{\sqrt{2}}^{\sqrt{3}}\frac{dt}{t^2-1}dt$$$$-2(arctan\left(\sqrt{3}\right)-arctan\left(\sqrt{2}\right))$$$$=2(arctan\left(\sqrt{3}\right)-arctan\left(\sqrt{2}\right))+ln\mid \frac{\sqrt{3}-1}{\sqrt{3}+1}\mid -ln\mid \frac{\sqrt{2}-1}{\sqrt{2}+1}\mid .$$$$$$

Commented by abdo imad last updated on 21/Feb/18

$$forgive...theQisfind{\int }_2^3\frac{\sqrt{x+1}}{x\sqrt{x-1}}dx.$$

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