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Question Number 91421 by  M±th+et+s last updated on 30/Apr/20

$${\int }_1^3\frac{1}{x\sqrt{3x^2+2x-1}}dx$$

Commented by abdomathmax last updated on 30/Apr/20

$$I={\int }_1^3\frac{dx}{x\sqrt{3x^2+2x-1}}$$$$3x^2+2x-1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1+3=4\Rightarrow x_1=\frac{-1+2}{3}=\frac{1}{3}$$$$x_2=\frac{-1-2}{3}=-1\Rightarrow I={\int }_1^3\frac{dx}{x\sqrt{3(x-\frac{1}{3})(x+1)}}$$$$={\int }_1^3\frac{dx}{x\sqrt{(3x-1)(x+1)}}={\int }_1^3\frac{dx}{x\sqrt{x+1}\sqrt{3x-1}}$$$$wedothechangement\sqrt{x+1}=t\Rightarrow x+1=t^2\Rightarrow$$$$I={\int }_{\sqrt{2}}^2\frac{2tdt}{(t^2-1)\sqrt{3(t^2-1)-1}}$$$$={\int }_{\sqrt{2}}^2\frac{2tdt}{(t^2-1)\sqrt{3t^2-4}}=\frac{2}{\sqrt{3}}{\int }_{\sqrt{2}}^2\frac{tdt}{(t^2-1)\sqrt{t^2-\frac{4}{3}}}$$$${=}_{t=\frac{2}{\sqrt{3}}ch\left(u\right)\to u=argch\left(\frac{t\sqrt{3}}{2}\right)}\frac{2}{\sqrt{3}}{\int }_{argch\left(\frac{\sqrt{6}}{2}\right)}^{argch\left(\sqrt{3}\right)}{\left(\frac{2}{\sqrt{3}}\right)}^2\times \frac{chushu}{(\frac{4}{\sqrt{3}}{ch}^{2}u-1)\frac{2}{\sqrt{3}}shu}du$$$$=\frac{4}{\sqrt{3}}{\int }_{argch\left(\frac{\sqrt{6}}{2}\right)}^{argch\left(\sqrt{3}\right)}\frac{shu}{(4{ch}^{2}u-\sqrt{3})}du$$$$=\frac{4}{\sqrt{3}}{\int }_{argch\left(\frac{\sqrt{6}}{2}\right)}^{argch\left(\sqrt{3}\right)}\frac{shu}{(4\times \frac{1+ch\left(2u\right)}{2}-\sqrt{3})}du$$$$=\frac{4}{\sqrt{3}}{\int }_{argch\left(\frac{\sqrt{6}}{2}\right)}^{argch\left(\sqrt{3}\right)}\frac{shu}{(2ch\left(2u\right)+2-\sqrt{3})}du$$$$=\frac{4}{\sqrt{3}}{\int }_{argch\left(\frac{\sqrt{6}}{2}\right)}^{argch\left(\sqrt{3}\right)}\frac{\frac{e^u-e^{-u}}{2}}{e^{2u}+e^{-2u}+2-\sqrt{3}}du$$$$=\frac{2}{\sqrt{3}}{\int }_{\frac{\sqrt{6}}{2}+\sqrt{\frac{6}{4}-1}}^{\sqrt{3}+\sqrt{2}}\frac{z-z^{-1}}{z^2+z^{-2}+2-\sqrt{3}}\frac{dz}{z}(z=e^u)$$$$=\frac{2}{\sqrt{3}}{\int }_{\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}}^{\sqrt{3}+\sqrt{2}}\frac{z^2-1}{z^4+1+(2-\sqrt{3})z^2}dz$$$$=\frac{2}{\sqrt{3}}{\int }_{\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}}^{\sqrt{3}+\sqrt{2}}\frac{z^2-1}{z^4+(2-\sqrt{3})z^2+1}dxletddcompose$$$$F\left(z\right)=\frac{z^2-1}{z^4+(2-\sqrt{3})z^2+1}...becontinued...$$$$$$$$$$

Commented by jagoll last updated on 30/Apr/20

$$\int \frac{dx}{x\sqrt{x^2(3+\frac{2}{x}-\frac{1}{x^2})}}$$$$\int \frac{dx}{x^2\sqrt{-(\frac{1}{x^2}-\frac{2}{x}+1)+4}}$$$$\int \frac{dx}{x^2\sqrt{4-{(\frac{1}{x}-1)}^2}}$$$$[let\frac{1}{x}-1=t$$$$\frac{dx}{x^2}=-dt$$$$\int \frac{-dt}{\sqrt{4-t^2}}.noweasytosolve$$

Answered by jagoll last updated on 30/Apr/20

Commented by  M±th+et+s last updated on 30/Apr/20

$$niceworkthanks$$

Answered by MJS last updated on 01/May/20

$$x^2(3x^2+2x-1)=x^2(x+1)(3x-1)$$$$\mathrm{let}x=\frac{1}{at+b}$$$$-\frac{(at+b-3)(at+b+1)}{{(at+b)}^4}$$$$at+b-3=t-c\iff (a-1)t+b+c-3=0$$$$at+b+1=t+c\iff (a-1)t+b-c+1=0$$$$\Rightarrow a=1$$$$b+c-3=0\wedge b-c+1=0\Rightarrow b=1\wedge k=2$$$$-\frac{(t-2)(t+2)}{{(t+1)}^4}$$$$\mathrm{let}t=2u$$$$-\frac{4(u-1)(u+1)}{{(2u+1)}^4}$$$$\Rightarrow$$$$\mathrm{let}x=\frac{1}{2\sin u+1}\iff u=\arcsin \frac{x-1}{2x}\to dx=x\sqrt{3x^2+2x-1}du$$$$\int \frac{dx}{x\sqrt{3x^2+2x-1}}=\int du=u=\arcsin \frac{x-1}{2x}+C$$

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