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Question Number 73144 by mathmax by abdo last updated on 06/Nov/19

$$calculte\int \frac{x+\sqrt{2+x^2}}{x+1-\sqrt{2+x^2}}dx$$

Commented by mathmax by abdo last updated on 07/Nov/19

$$I=\int \frac{x+\sqrt{2+x^2}}{x+1-\sqrt{2+x^2}}dxchangementx=\sqrt{2}sh\left(t\right)give$$$$I=\int \frac{\sqrt{2}sh\left(t\right)+\sqrt{2}ch\left(t\right)}{\sqrt{2}sh\left(t\right)+1-\sqrt{2}cht}\sqrt{2}chtdt$$$$=\int \frac{(2sh\left(t\right)ch\left(t\right)+{ch}^{2}t)}{\sqrt{2}sh\left(t\right)-\sqrt{2}ch\left(t\right)+1}dt=\int \frac{sh\left(2t\right)+\frac{1+ch\left(2t\right)}{2}}{\sqrt{2}sh\left(t\right)-\sqrt{2}ch\left(t\right)+1}dt$$$$=\int \frac{2sh\left(2t\right)+ch\left(2t\right)+1}{2(\sqrt{2}sh\left(t\right)-\sqrt{2}ch\left(t\right)+1)}dt$$$$=\frac{1}{2}\int \frac{2\frac{e^{2t}-e^{-2t}}{2}+\frac{e^{2t}+e^{-2t}}{2}+1}{\sqrt{2}\frac{e^t-e^{-t}}{2}-\sqrt{2}\frac{e^t+e^{-t}}{2}+1}dt$$$$=\frac{1}{2}\int \frac{2e^{2t}-2e^{-2t}+e^{2t}+e^{-2t}+2}{\sqrt{2}{e}^t-\sqrt{2}{e}^{-t}-\sqrt{2}{e}^t-\sqrt{2}{e}^{-t}+2}dt\Rightarrow$$$$2I=\int \frac{3e^{2t}-e^{-2t}+2}{-2\sqrt{2}{e}^{-t}+2}dt{=}_{e^t=u}\int \frac{3u^2-u^{-2}+2}{2-2\sqrt{2}{u}^{-1}}\frac{du}{u}$$$$=\int \frac{3u^2-u^{-2}}{2u-2\sqrt{2}}du=\int \frac{3u^4-1}{2u^3-2\sqrt{2}u}du\Rightarrow$$$$4I=\int \frac{3u^4-1}{u^3-\sqrt{2}u}du=\int \frac{3u(u^3-\sqrt{2}u)+3\sqrt{2}{u}^2-1}{u^3-\sqrt{2}u}du$$$$=3u+\int \frac{3\sqrt{2}{u}^2-1}{u(u^2-\sqrt{2})}duletdecomposeF\left(u\right)=\frac{3\sqrt{2}{u}^2-1}{u(u^2-\sqrt{2})}$$$$\Rightarrow F\left(u\right)=\frac{3\sqrt{2}{u}^2-1}{u(u-\alpha )(u+\alpha )}with\alpha {=}^4\sqrt{2}$$$$F\left(u\right)=\frac{a}{u}+\frac{b}{u-\alpha }+\frac{c}{u+\alpha }$$$$a=\frac{-1}{-{\alpha }^2}=\frac{1}{{\alpha }^2}=\frac{1}{\sqrt{2}}$$$$b=\frac{3\sqrt{2}{\alpha }^2-1}{\alpha \left(2\alpha \right)}=\frac{3\sqrt{2}{\alpha }^2-1}{2{\alpha }^2}andc=\frac{3\sqrt{2}{\alpha }^2-1}{(-\alpha )(-2\alpha )}=\frac{3\sqrt{2}{\alpha }^2-1}{2{\alpha }^2}and$$$$\int F\left(u\right)du=aln\mid u\mid +bln\mid u-\alpha \mid +cln\mid u+\alpha \mid c$$$$=at+bln\mid e^t-\alpha \mid +cln\mid e^t+\alpha \mid +cbutt=argsh\left(\frac{x}{\sqrt{2}}\right)$$$$=ln(\frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}})\Rightarrow$$$$\int F\left(u\right)du=aln(\frac{x}{2}+\sqrt{1+\frac{x^2}{2}})+bln\mid \frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}}-\alpha \mid$$$$+cln\mid \frac{x}{\sqrt{2}}+\sqrt{1+\frac{x^2}{2}}+\alpha \mid +csothevalueofIisknown.$$$$$$$$$$

Answered by MJS last updated on 06/Nov/19

$$2\mathrm{substitutions}\mathrm{in}1\mathrm{step}$$$$x=\sqrt{2}\sinh s\wedge s=\ln t$$$$\Rightarrow$$$$x=\frac{t^2-1}{t\sqrt{2}}\iff t=\frac{x+\sqrt{x^2+2}}{\sqrt{2}}\to dx=\frac{\sqrt{2(x^2+2)}}{x+\sqrt{x^2+2}}dt=\frac{t^2+1}{t^2\sqrt{2}}dt$$$$\int \frac{x+\sqrt{x^2+2}}{x+1-\sqrt{x^2+2}}dx=\int \frac{t^2+1}{t-\sqrt{2}}dt=$$$$=\int tdt+\sqrt{2}\int dt+3\int \frac{dt}{t-\sqrt{2}}=$$$$=\frac{1}{2}{t}^2+t\sqrt{2}+3\ln (t-\sqrt{2})=$$$$=\frac{(x+2)(x+\sqrt{x^2+2})}{2}+3\ln (x-2+\sqrt{x^2+2})+C$$

Commented by mathmax by abdo last updated on 07/Nov/19

$$thankyousirmjs.$$

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