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Question Number 109839 by john santu last updated on 25/Aug/20

$$\frac{JS}{\approx \mathrm{♡}\approx }$$$$\left(1\right)\int \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}dx$$$$\left(2\right)\int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$$

Commented by Her_Majesty last updated on 25/Aug/20

$$\left(1\right)=\int x-\sqrt{x^2-1}dx$$$$\left(2\right)lett=\sqrt{tanx}$$$$bothare\mathrm{easiest}$$

Answered by Dwaipayan Shikari last updated on 25/Aug/20

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

Answered by bobhans last updated on 26/Aug/20

$$(\iff )\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}=\frac{{(\sqrt{x+1}-\sqrt{x-1})}^2}{(x+1)-(x-1)}$$$$=\frac{x+1+x-1-2\sqrt{x^2-1}}{2}=\frac{2x-2\sqrt{x^2-1}}{2}$$$$=x-\sqrt{x^2-1}$$$$nowI=\int (x-\sqrt{x^2-1})dx=\frac{x^2}{2}-\int \sqrt{x^2-1}dx$$$$letx=\sec s\Rightarrow dx=\sec s\tan sds$$$$I=\frac{x^2}{2}-\int \sec s\tan {}^{2}sds$$$$I=\frac{x^2}{2}-\int \sec s(\sec {}^{2}s-1)ds$$$$I=\frac{x^2}{2}+\ln \mid \sec s+\tan s\mid -\int \sec {}^{3}sds$$$$let{I}_2=\int \sec {}^{3}sds=\int \sec sd\left(\tan s\right)$$$$I_2=\sec s\tan s-\int \sec s\tan {}^{2}sds$$$$I_2=\sec s\tan s-\int (\sec {}^{3}s-\sec s)ds$$$$2I_2=\sec s\tan s+\ln \mid \sec s+\tan s\mid$$$$I_2=\frac{1}{2}\sec s\tan s+\frac{1}{2}\ln \mid \sec s+\tan s\mid$$$$weconcludeI=\frac{x^2}{2}+\frac{1}{2}\ln \mid \sec s+\tan s\mid -\frac{1}{2}\sec s\tan s+c$$$$I=\frac{x^2+\ln \mid x+\sqrt{x^2-1}\mid -x\sqrt{x^2-1}+C}{2}$$

Answered by bobhans last updated on 26/Aug/20

$$\left(2\right)set\tan x=z^2\Rightarrow \sec {}^{2}xdx=2zdz$$$$\Rightarrow dx=\frac{2zdz}{1+z^4}$$$$I=\int \frac{z}{1+z}\times \frac{2zdz}{1+z^4}=\int \frac{2z^2}{(1+z)(1+z^4)}dz$$$$\frac{2z^2}{(1+z)(1+z^4)}=\frac{A}{1+z}+\frac{{Bz}^3+{Cz}^2+Dz+E}{1+z^4}$$$$2z^2=A(1+z^4)+({Bz}^3+{Cz}^2+Dz+E)(1+z)$$$$z=0\Rightarrow 0=A+E$$$$z=-1\Rightarrow 2=2A,A=1\wedge E=-1$$$$z=1\Rightarrow 2=2+(B+C+D-1).2$$$$0=(B+C+D-1)2\to B+C+D=1$$$$continue$$

Commented by Sarah85 last updated on 26/Aug/20

$$\mathrm{you}\mathrm{cannot}\mathrm{set}\mathrm{up}\frac{Bz+C}{1+z^4}$$

Commented by bobhans last updated on 26/Aug/20

$$ooyesitshouldbe\frac{{Bz}^3+{Cz}^2+Dz+E}{1+z^4}$$

Answered by Sarah85 last updated on 26/Aug/20

$$\int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$$$$t=\sqrt{tanx}\mathrm{leads}\mathrm{to}$$$$2\int \frac{t^2}{(t+1)(t^4+1)}dt=$$$$=\int \frac{dt}{t+1}-\int \frac{{(t-1)}^2(t+1)}{(t^4+1)}dt=$$$$=\int \frac{dt}{t+1}-\frac{1-\sqrt{2}}{2}\int \frac{t+1}{t^2-\sqrt{2}t+1}dt-\frac{1+\sqrt{2}}{2}\int \frac{t+1}{t^2+\sqrt{2}t+1}dt$$$$\int \frac{dt}{t+1}=\ln (t+1)$$$$-\frac{1-\sqrt{2}}{2}\int \frac{t+1}{t^2-\sqrt{2}t+1}dt=-\frac{1-\sqrt{2}}{4}\ln (t^2-\sqrt{2}t+1)+\frac{1}{2}{\tan }^{-1}(\sqrt{2}t-1)$$$$-\frac{1+\sqrt{2}}{2}\int \frac{t+1}{t^2+\sqrt{2}t+1}dt=-\frac{1+\sqrt{2}}{4}\ln (t^2+\sqrt{2}t+1)-\frac{1}{2}{\tan }^{-1}(\sqrt{2}t+1)$$$$\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{complete}$$

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