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Question Number 69792 by mathmax by abdo last updated on 27/Sep/19

$$findf\left(\alpha \right)=\int \frac{dx}{x+\alpha +\sqrt{x^2+3}}$$$$andg\left(\alpha \right)=\int \frac{dx}{{(x+\alpha +\sqrt{x^2+3})}^2}with\alpha real$$

Commented by mathmax by abdo last updated on 10/Oct/19

$$f\left(\alpha \right)=\int \frac{dx}{x+\alpha +\sqrt{x^2+3}}changementx=\sqrt{3}sh\left(t\right)give$$$$f\left(\alpha \right)=\int \frac{\sqrt{3}ch\left(t\right)dt}{\sqrt{3}sh\left(t\right)+\alpha +\sqrt{3}ch\left(t\right)}=\int \frac{\frac{e^t+e^{-t}}{2}}{\frac{e^t-e^{-t}}{2}+\alpha +\frac{e^t+e^{-t}}{2}}dt$$$$=\int \frac{e^t+e^{-t}}{e^t-e^{-t}+2\alpha +e^t+e^{-t}}dt=\int \frac{e^t+e^{-t}}{2e^t+2\alpha }dt$$$${=}_{e^t=u}\frac{1}{2}\int \frac{u+u^{-1}}{u+\alpha }\frac{du}{u}=\frac{1}{2}\int \frac{u^2+1}{u^2(u+\alpha )}duletdecompose$$$$F\left(u\right)=\frac{u^2+1}{u^2(u+\alpha )}\Rightarrow F\left(u\right)=\frac{a}{u}+\frac{b}{u+\alpha }+\frac{c}{u^2}$$$$b={lim}_{u\to -\alpha }(u+\alpha )F\left(u\right)=\frac{1+{\alpha }^2}{{\alpha }^2}$$$$c={lim}_{u\to 0}{u}^{2}F\left(u\right)=\frac{1}{\alpha }\Rightarrow F\left(u\right)=\frac{a}{u}+\frac{1+{\alpha }^2}{{\alpha }^2(u+\alpha )}+\frac{1}{\alpha u^2}$$$${lim}_{u\to +\mathrm{\infty }}uF\left(u\right)=1=a+\frac{1+{\alpha }^2}{{\alpha }^2}\Rightarrow a=1-\frac{1}{{\alpha }^2}-1=-\frac{1}{{\alpha }^2}\Rightarrow$$$$F\left(u\right)=-\frac{1}{{\alpha }^{2}u}+\frac{1+{\alpha }^2}{{\alpha }^2(u+\alpha )}+\frac{1}{\alpha u^2}\Rightarrow$$$$\int F\left(u\right)du=-\frac{1}{{\alpha }^2}ln\mid u\mid +\frac{1+{\alpha }^2}{{\alpha }^2}ln\mid u+\alpha \mid -\frac{1}{\alpha u}+c$$$$=-\frac{t}{{\alpha }^2}+\frac{1+{\alpha }^2}{{\alpha }^2}ln\mid e^t+\alpha \mid -\frac{1}{\alpha }{e}^{-t}+cbutt=argsh\left(\frac{x}{\sqrt{3}}\right)$$$$=ln(\frac{x}{\sqrt{3}}+\sqrt{1+\frac{x^2}{3}})\Rightarrow$$$$\int F\left(u\right)du=-\frac{1}{{\alpha }^2}ln(\frac{x}{\sqrt{3}}+\sqrt{1+\frac{x^2}{3}})+\frac{1+{\alpha }^2}{{\alpha }^2}ln\mid \frac{x}{3}+\sqrt{1+\frac{x^2}{3}}+\alpha \mid$$$$-\frac{1}{\alpha (\frac{x}{3}+\sqrt{1+\frac{x^2}{3}})}+C\Rightarrow$$$$f\left(\alpha \right)=-\frac{1}{2{\alpha }^2}ln(\frac{x}{\sqrt{3}}+\sqrt{1+\frac{x^2}{3}})+\frac{1+{\alpha }^2}{2{\alpha }^2}ln\mid \frac{x}{3}+\sqrt{1+\frac{x^2}{3}}+\alpha \mid$$$$-\frac{1}{2\alpha (\frac{x}{3}+\sqrt{1+\frac{x^2}{3}})}+C.$$

Answered by MJS last updated on 28/Sep/19

$$\int \frac{dx}{x+\alpha +\sqrt{x^2+3}}=$$$$[\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{substitute}x=\sqrt{3}\sinh t\mathrm{and}$$$$\mathrm{then}t=\ln u$$$$\Rightarrow \mathrm{both}\mathrm{in}1\mathrm{step}\mathrm{gives}$$$$u=\frac{\sqrt{3}}{3}(x+\sqrt{x^2+3})\to dx=\frac{\sqrt{3}(u^2+1)}{2u^2}du$$$$x=\frac{\sqrt{3}(u^2-1)}{2u}]$$$$=\frac{\sqrt{3}}{2}\int \frac{u^2+1}{u^2(\sqrt{3}u+\alpha )}du=$$$$=\frac{\sqrt{3}}{2\alpha }\int \frac{du}{u^2}-\frac{3}{2{\alpha }^2}\int \frac{du}{u}+\frac{\sqrt{3}({\alpha }^2+3)}{2{\alpha }^2}\int \frac{du}{\sqrt{3}u+\alpha }=$$$$=-\frac{\sqrt{3}}{2\alpha }{u}^{-1}-\frac{3}{2{\alpha }^2}\ln u+\frac{{\alpha }^2+3}{2{\alpha }^2}\ln (\sqrt{3}u+\alpha )=$$$$=\frac{1}{2\alpha }(x-\sqrt{x^2+3})-\frac{3}{2{\alpha }^2}\ln (x+\sqrt{x^2+3})+\frac{{\alpha }^2+3}{2{\alpha }^2}\ln (x+\alpha +\sqrt{x^2+3})+C$$$$$$$$\int \frac{dx}{{(x+\alpha +\sqrt{x^2+3})}^2}=$$$$\left[\mathrm{same}\mathrm{substitution}\mathrm{as}\mathrm{above}\right]$$$$=\frac{\sqrt{3}}{2}\int \frac{u^2+1}{u^2{(\sqrt{3}u+\alpha )}^2}du=$$$$=\frac{\sqrt{3}}{2{\alpha }^2}\int \frac{du}{u^2}-\frac{3}{{\alpha }^3}\int \frac{du}{u}+\frac{\sqrt{3}({\alpha }^2+3)}{2{\alpha }^2}\int \frac{du}{{(\sqrt{3}u+\alpha )}^2}+\frac{3\sqrt{3}}{{\alpha }^3}\int \frac{du}{\sqrt{3}u+\alpha }=$$$$=-\frac{\sqrt{3}}{2{\alpha }^2}{u}^{-1}-\frac{3}{{\alpha }^3}\ln u-\frac{{\alpha }^2+3}{2{\alpha }^2(\sqrt{3}u+\alpha )}+\frac{3}{{\alpha }^3}\ln (\sqrt{3}u+\alpha )=$$$$=\frac{1}{2{\alpha }^2}(x-\sqrt{x^2+3})-\frac{3}{{\alpha }^3}\ln (x+\sqrt{x^2+3})-\frac{{\alpha }^2+3}{2{\alpha }^2(x+\alpha +\sqrt{x^2+3})}+\frac{3}{{\alpha }^3}\ln (x+\alpha +\sqrt{x^2+3})+C=$$$$=\frac{2\alpha x^2-6x-\alpha ({\alpha }^2+3)-2(\alpha x-3)\sqrt{x^2+3}}{2{\alpha }^2(2\alpha x+{\alpha }^2-3)}+\frac{3}{{\alpha }^3}\ln (-\alpha x+3+\alpha \sqrt{x^2+3})+C$$

Commented by mathmax by abdo last updated on 28/Sep/19

$$thankyousirmjsforthishardwork.$$

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