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Question Number 177320 by mnjuly1970 last updated on 03/Oct/22

Answered by a.lgnaoui last updated on 04/Oct/22

$$\sqrt{2}+(1+\sqrt{2})x+x^2={(x+\frac{1+\sqrt{2}}{2})}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2=(x+\frac{1+\sqrt{2}}{2}+\frac{\sqrt{3}}{2})(x+\frac{1+\sqrt{2}}{2}-\frac{\sqrt{3}}{2})$$$$\frac{1}{(x+\frac{1+\sqrt{2}}{2}-\frac{\sqrt{3}}{2})(x+\frac{1+\sqrt{2}}{2}+\frac{\sqrt{3}}{2})}=-\frac{2\sqrt{3}}{3}[\frac{1}{(x+\frac{1+\sqrt{2}-\sqrt{3}}{2})}+\frac{1}{(x+\frac{1+\sqrt{2}+\sqrt{3}}{2})}]$$$${\int }_0^{\mathrm{\infty }}\frac{\ln x}{x^2+(1+\sqrt{2})x}dx=-\frac{2\sqrt{3}}{3}\int (\frac{1}{x+\frac{1+\sqrt{2}-\sqrt{3}}{2}}+\frac{1}{x+\frac{1+\sqrt{3}+\sqrt{3}}{2}})\ln xdx$$$$=\ln {\left[x(x+\frac{1+\sqrt{2}-\sqrt{3}}{2})(x+\frac{1+\sqrt{2}+\sqrt{3}}{2})\right]}_0^{\mathrm{\infty }}+\frac{2\sqrt{3}}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{x}\left(\frac{1}{x+\frac{1+\sqrt{2}-\sqrt{3}}{2}}\right)+\left(\frac{1}{x+\frac{1+\sqrt{2}+\sqrt{3}}{2}}\right)dx$$$$\frac{1}{x(x+\frac{1+\sqrt{2}-\sqrt{3}}{2})}=\frac{1}{{(x+\frac{1+\sqrt{2}-\sqrt{3}}{4})}^2-\left(\frac{3-(\sqrt{2}-\sqrt{3}-\sqrt{6}}{8}\right)}dememepour\frac{1}{x(x+\frac{1+\sqrt{2^{}}+\sqrt{3}}{2})}$$$$onaalirsforme\frac{1}{{\mathrm{x}}^2-{\mathrm{a}}^2}=\frac{1}{{\mathrm{a}}^2[{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^2-1]}$$$$=\frac{2}{\sqrt{\sqrt{6}+\sqrt{3}-\sqrt{2}+3}}{\left[Arc\tan (x+\frac{1+\sqrt{2}+\sqrt{3}}{4})\right]}_0^{\mathrm{\infty }}=\frac{\pi }{2}-\left(\frac{8}{\sqrt{3}+\sqrt{6}+\sqrt{2}+3}\right)arc\tan \left(\frac{17+\sqrt{2^{}}+\sqrt{3}}{4}\right)$$$$...................$$$$$$$$$$$$$$$$$$$$$$$$$$

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