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Question Number 41487 by behi83417@gmail.com last updated on 08/Aug/18

Commented by maxmathsup by imad last updated on 08/Aug/18

$$2)wehaveI=\int \frac{dx}{(1+x)\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}changementx+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)give$$$$I=\int \frac{1}{(1+\frac{\sqrt{3}}{2}sh\left(t\right)-\frac{1}{2})\frac{\sqrt{3}}{2}ch\left(t\right)}\frac{\sqrt{3}}{2}ch\left(t\right)dt$$$$=\int \frac{dt}{\frac{1}{2}+\frac{\sqrt{3}}{2}sh\left(t\right)}=\int \frac{2dt}{\sqrt{3}\frac{e^t-e^{-t}}{2}+\frac{1}{2}}=\int \frac{4dt}{\sqrt{3}{e}^t-\sqrt{3}{e}^{-t}+1}$$$${=}_{e^t=u}\int \frac{4}{\sqrt{3}u-\sqrt{3}\frac{1}{u}+1}\frac{du}{u}=\int \frac{4du}{\sqrt{3}{u}^2-\sqrt{3}+u}letdecompose$$$$F\left(u\right)=\frac{4}{\sqrt{3}{u}^2+u-\sqrt{3}}wehave\mathrm{\Delta }=1-4(-3)=13\Rightarrow$$$$u_1=\frac{-1+\sqrt{13}}{2\sqrt{3}}and{u}_2=\frac{-1-\sqrt{13}}{2\sqrt{3}}\Rightarrow F\left(u\right)=\frac{4}{\sqrt{3}(u-u_1)(u-u_2)}$$$$=\frac{a}{u-u_1}+\frac{b}{u-u_2}$$$$a=\frac{4}{\sqrt{3}(u_1-u_2)}=\frac{4}{\sqrt{3}\frac{\sqrt{13}}{\sqrt{3}}}=\frac{4}{\sqrt{3}}$$$$b=\frac{4}{\sqrt{3}(u_2-u_1)}=-\frac{4}{\sqrt{3}}\Rightarrow F\left(u\right)=\frac{4}{\sqrt{3}(u-u_1)}-\frac{4}{\sqrt{3}(u-u_2)}\Rightarrow$$$$\int F\left(u\right)du=\frac{4}{\sqrt{3}}ln\mid u-u_1\mid -\frac{4}{\sqrt{3}}ln\mid u-u_2\mid +c$$$$=\frac{4}{\sqrt{3}}ln\mid e^t-u_1\mid -\frac{4}{\sqrt{3}}ln\mid e^t-u_2\mid +cbutt=argsh\left(\frac{2x+1}{\sqrt{3}}\right)$$$$=ln(\frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}\Rightarrow$$$$I=\frac{4}{\sqrt{3}}ln\mid \frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{2}\right)}^2}-u_1\mid -\frac{4}{\sqrt{3}}ln\mid \frac{2x+1}{\sqrt{3}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}-u_2\mid +c$$$$$$

Commented by maxmathsup by imad last updated on 08/Aug/18

$$youarewelcomesir$$

Commented by behi83417@gmail.com last updated on 08/Aug/18

$$thankyoudearproph.abdo.$$

Commented by math khazana by abdo last updated on 10/Aug/18

$$1)letusethechangement\sqrt{\frac{1+x^2}{1-x^2}}=t\Rightarrow$$$$\frac{1+x^2}{1-x^2}=t^2\Rightarrow 1+x^2=t^2-t^2{x}^2\Rightarrow (1+t^2)x^2=t^2-1\Rightarrow$$$$x^2=\frac{t^2-1}{t^2+1}\Rightarrow 2xdx=\frac{2t(t^2+1)-2t(t^2-1)}{{(t^2+1)}^2}$$$$=\frac{4t}{{(t^2+1)}^2}dt\Rightarrow$$$$\int \frac{1}{x}\sqrt{\frac{1+x^2}{1-x^2}}dx=\int \frac{1}{2x^2}\sqrt{\frac{1+x^2}{1-x^2}}\left(2x\right)dx$$$$=\frac{1}{2}\int \frac{t^2+1}{t^2-1}.t.\frac{4t}{{(t^2+1)}^2}dt$$$$=\int \frac{2t^2}{(t^2-1)(t^2+1)}dtletdecompose$$$$F\left(t\right)=\frac{2t^2}{(t^2-1)(t^2+1)}=\frac{2t^2}{(t-1)(t+1)(t^2+1)}$$$$F\left(t\right)=\frac{a}{t-1}+\frac{b}{t+1}+\frac{ct+d}{t^2+1}$$$$a={lim}_{t\to 1}(t-1)F\left(t\right)=\frac{2}{4}=\frac{1}{2}$$$$b={lim}_{t\to -1}(t+1)F\left(t\right)=\frac{2}{-4}=-\frac{1}{2}\Rightarrow$$$$F\left(t\right)=\frac{1}{2(t-1)}-\frac{1}{2(t+1)}+\frac{ct+d}{t^2+1}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=c\Rightarrow$$$$F\left(t\right)=\frac{1}{2(t-1)}-\frac{1}{2(t+1)}+\frac{d}{t^2+1}$$$$F\left(0\right)=-\frac{1}{2}-\frac{1}{2}+d=0\Rightarrow d=1\Rightarrow$$$$F\left(t\right)=\frac{1}{2(t-1)}-\frac{1}{2(t+1)}+\frac{1}{t^2+1}$$$$I=\int F\left(t\right)dt=\frac{1}{2}ln\mid \frac{t-1}{t+1}\mid +arctan\left(t\right)+c$$$$=\frac{1}{2}ln\mid \frac{\sqrt{\frac{1+x^2}{1-x^2}}-1}{\sqrt{\frac{1+x^2}{1-x^2}}+1}\mid +arctan\left(\sqrt{\frac{1+x^2}{1-x^2}}\right)+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

$$2)1+x=\frac{1}{t}dx=-\frac{1}{t^2}dt$$$$\int \frac{-dt}{t^2\times \frac{1}{t}\times \sqrt{1+(\frac{1}{t}-1)+{(\frac{1}{t}-1)}^2}}$$$$\int \frac{-dt}{t\times \sqrt{\frac{1}{t}+\frac{1}{t^2}-\frac{2}{t}+1}}$$$$\int \frac{-dt}{t\times \sqrt{\frac{t+1-2t+t^2}{t^2}}}$$$$\int \frac{-dt}{\sqrt{t^2-t+1}}$$$$\int \frac{-dt}{\sqrt{t^2-2.t.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}}$$$$\int \frac{-dt}{\sqrt{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$=-ln\{(t-\frac{1}{2})+\sqrt{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$=-ln\{(1+x-\frac{1}{2})+\sqrt{{(1+x-\frac{1}{2})}^2+\frac{3}{4}}$$$$=-ln\{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+c$$$$$$$$$$

Commented by behi83417@gmail.com last updated on 08/Aug/18

$$thankyouverymuchdeartanmaysir.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

$$thankyousir...forkeepingourmindbusy...$$$$$$

Commented by behi83417@gmail.com last updated on 08/Aug/18

$$thanks.Ienjoytoseedifferent$$$$questions,iedas,andsolutions.$$$$Ialwyeshavegoodsenseandtime$$$$whenbeinginthisforum.$$$$thankstoallmybestfriends$$$$andcolleagueshere.$$$$Godblessallofyou.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18

$$1)\int \frac{1}{x}.\sqrt{\frac{1+x^2}{1-x^2}}dx$$$$\int \frac{1}{x}.\sqrt{\frac{1+x^2}{1-x^2}}dx$$$$t^2=\frac{1+x^2}{1-x^2}$$$$1+x^2=t^2-t^2{x}^2$$$$x^2(1+t^2)=t^2-1$$$$x^2=\frac{t^2-1}{t^2+1}$$$$2xdx=\frac{(t^2+1)2t-(t^2-1)\left(2t\right)}{{(t^2+1)}^2}dt$$$$2xdx=\frac{2t^3+2t-2t^3+2t}{{(t^2+1)}^2}dt$$$$xdx=\frac{2tdt}{{(t^2+1)}^2}$$$$\int \frac{1}{x}.\sqrt{\frac{1+x^2}{1-x^2}}dx$$$$\int \sqrt{\frac{1+x^2}{1-x^2}}.\frac{xdx}{x^2}$$$$\int t.\frac{2tdt}{{(t^2+1)}^2}.\frac{t^2+1}{t^2-1}$$$$\int \frac{2t^2}{(t^2+1)(t^2-1)}dt$$$$\int \frac{t^2+1+t^2-1}{(t^2+1)(t^2-1)}dt$$$$\int \frac{dt}{t^2-1}+\int \frac{dt}{t^2+1}$$$$=\frac{1}{2}\int \frac{(t+1)-(t-1)}{(t+1)(t-1)}dt+\int \frac{dt}{t^2+1}$$$$=\frac{1}{2}[\int \frac{dt}{t-1}-\int \frac{dt}{t+1}]-\int \frac{dt}{t^2+1}$$$$=\frac{1}{2}ln\left(\frac{t-1}{t+1}\right)-{tan}^{-1}\left(t\right)+c$$$$=\frac{1}{2}ln\left(\frac{\sqrt{\frac{1+x^2}{1-x^2}}-1}{\sqrt{\frac{1+x^2}{1-x^2}}+1}\right)-{tan}^{-1}\left(\sqrt{\frac{1+x^2}{1-x^2}}\right)+c$$$$=\frac{1}{2}ln\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)-{tan}^{-1}\left(\sqrt{\frac{1+x^2}{1-x^2}}\right)+c$$

Commented by behi83417@gmail.com last updated on 08/Aug/18

$$nicework.thanksinadvancesir.$$

Commented by behi83417@gmail.com last updated on 09/Aug/18

$$=\frac{1}{2}ln\left(\frac{1-\sqrt{1-x^4}}{x^2}\right)-{tg}^{-1}\left(\frac{1+x^2}{\sqrt{1-x^4}}\right)+c$$

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

$$thankyousir...$$

Answered by Ar Brandon last updated on 28/Jul/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{(1+\mathrm{x})\sqrt{1+\mathrm{x}+{\mathrm{x}}^2}}=\int \frac{\mathrm{dx}}{(\mathrm{x}+\frac{1}{2}+\frac{1}{2})\sqrt{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}}$$$$\mathrm{x}+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan \theta \Rightarrow \mathrm{dx}=\frac{\sqrt{3}}{2}{\sec }^2\theta \mathrm{d}\theta$$$$\Rightarrow \mathcal{I}=\int \frac{\frac{\sqrt{3}}{2}{\sec }^2\theta \mathrm{d}\theta }{(\frac{\sqrt{3}}{2}\tan \theta +\frac{1}{2})\sqrt{\frac{3}{4}{\tan }^2\theta +\frac{3}{4}}}$$$$=\int \frac{\sec \theta \mathrm{d}\theta }{\frac{\sqrt{3}}{2}\tan \theta +\frac{1}{2}}=\int \frac{\mathrm{d}\theta }{\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta }$$$$=\int \frac{\mathrm{d}\theta }{\sin \frac{\pi }{3}\sin \theta +\cos \frac{\pi }{3}\cos \theta }=\int \frac{\mathrm{d}\theta }{\cos (\theta -\frac{\pi }{3})}$$$$=\int \sec (\theta -\frac{\pi }{3})\mathrm{d}\theta =\ln \mid \sec (\theta -\frac{\pi }{3})+\tan (\theta -\frac{\pi }{3})\mid +\mathrm{lnA}$$

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