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Question Number 167823 by cortano1 last updated on 26/Mar/22

Commented by dangduomg last updated on 26/Mar/22

$$\mathrm{actually}\mathrm{very}\mathrm{long}\mathrm{answer}$$

Commented by MJS_new last updated on 26/Mar/22

$$\mathrm{a}\mathrm{first}\mathrm{steps}$$$$\int \frac{\ln (1+\sqrt{x^2-\frac{1}{3}})}{x\sqrt{x^2-\frac{1}{3}}}dx=$$$$[t=1+\sqrt{x^2-\frac{1}{3}}\to dx=\frac{\sqrt{x^2-\frac{1}{3}}}{x}dt]$$$$=\int \frac{\ln t}{t^2-2t+\frac{4}{3}}dt=$$$$=\int \frac{\ln t}{(t-1-\frac{\sqrt{3}}{3}\mathrm{i})(t-1+\frac{\sqrt{3}}{3}\mathrm{i})}dt=...$$$$\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{but}\mathrm{it}\mathrm{takes}\mathrm{some}\mathrm{time}...$$$$\frac{\ln t}{(t-a)(t-b)}=\frac{\ln t}{(a-b)(t-a)}+\frac{\ln t}{(b-a)(t-b)}$$$$...$$

Answered by Mathspace last updated on 27/Mar/22

$$f\left(a\right)=\int \frac{ln(1+a\sqrt{x^2-\frac{1}{3}})}{x\sqrt{x^2-\frac{1}{3}}}dx\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\int \frac{1}{x(1+a\sqrt{x^2-\frac{1}{3}})}dx$$$${=}_{x=\frac{1}{\sqrt{3}}cht}\int \frac{1}{\sqrt{3}.\frac{1}{\sqrt{3}}cht(1+a\frac{1}{\sqrt{3}}sht)}shtdt$$$$=\sqrt{3}\int \frac{sht}{cht(\sqrt{3}+asht)}dt$$$$=\sqrt{3}\int \frac{\frac{e^t-e^{-t}}{2}}{\frac{e^t+e^{-t}}{2}(\sqrt{3}+a\frac{e^t-e^{-t}}{2})}dt$$$$=2\sqrt{3}\int \frac{e^t-e^{-t}}{(e^t+e^{-t})(2\sqrt{3}+{ae}^t-{ae}^{-t})}dt$$$${=}_{e^t=z}\int \frac{z-z^{-1}}{(z+z^{-1})(2\sqrt{3}+az-{az}^{-1})}\frac{dz}{z}$$$$=\int \frac{z(z-z^{-1})}{z(z+z^{-1})z(2\sqrt{3}+az-{az}^{-1})}dz$$$$=\int \frac{z^2-1}{(z^2+1)(2\sqrt{3}z+{az}^2-a)}dz$$$$wedecompose$$$$F\left(z\right)=\frac{z^2-1}{(z^2+1)({az}^2+2\sqrt{3}z-a)}$$$$rootsof{az}^2+2\sqrt{3}z-a$$$${\mathrm{\Delta }}^{{}^{\prime }}=3+a^2\Rightarrow z_1=\frac{-\sqrt{3}+\sqrt{a^2+3}}{a}$$$$z_2=\frac{-\sqrt{3}-\sqrt{a^2+3}}{a}$$$$F\left(z\right)=\frac{\alpha z+\beta }{z^2+1}+\frac{c_1}{z-z_1}+\frac{c_2}{z-z_2}$$$$....becontinued...$$$$$$

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