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Question Number 82700 by jagoll last updated on 23/Feb/20

$$\int \frac{2dx}{3x\sqrt{5x^2+6}}?$$

Commented by john santu last updated on 23/Feb/20

$$\int \frac{d\left(x^2\right)}{3x^2\sqrt{5x^2+6}},let{x}^2=y$$$$\Rightarrow \int \frac{dy}{3y\sqrt{5y+6}},letagain\sqrt{5y+6}=t$$$$5y+6=t^2\Rightarrow dy=\frac{2}{5}tdt$$$$\frac{2}{15}\int \frac{tdt}{\left(\frac{t^2-6}{5}\right)t}=\frac{2}{3}\int \frac{dt}{t^2-{\left(\sqrt{6}\right)}^2}$$$$=\frac{2}{3}\frac{1}{2\sqrt{6}}ln\mid \frac{t-\sqrt{6}}{t+\sqrt{6}}\mid +c$$$$=\frac{1}{3\sqrt{6}}ln\mid \frac{\sqrt{5x^2+6}-\sqrt{6}}{\sqrt{5x^2+6}+\sqrt{6}}\mid +c$$

Commented by jagoll last updated on 23/Feb/20

$$wawcreatifsir.thankyou$$

Commented by mathmax by abdo last updated on 23/Feb/20

$$letI=\int \frac{2dx}{3x\sqrt{5x^2+6}}\Rightarrow \frac{3}{2}I=\int \frac{dx}{x\sqrt{5(x^2+\frac{6}{5})}}$$$${=}_{x=\sqrt{\frac{6}{5}}u}\frac{1}{\sqrt{5}}\int \frac{\sqrt{\frac{6}{5}}du}{\sqrt{\frac{6}{5}}u\times \sqrt{\frac{6}{5}}\sqrt{1+u^2}}=\frac{1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{6}}\int \frac{du}{u\sqrt{1+u^2}}$$$$=\frac{1}{\sqrt{6}}\int \frac{du}{u\sqrt{1+u^2}}{=}_{u=sh\left(z\right)}\frac{1}{\sqrt{6}}\int \frac{ch\left(z\right)}{sh\left(z\right)ch\left(z\right)}dz$$$$=\frac{1}{\sqrt{6}}\int \frac{dz}{\frac{e^z-e^{-z}}{2}}=\frac{2}{\sqrt{6}}\int \frac{dz}{e^z-e^{-z}}{=}_{e^z=\alpha }\frac{2}{\sqrt{6}}\int \frac{d\alpha }{\alpha (\alpha -{\alpha }^{-1})}$$$$=\frac{2}{\sqrt{6}}\int \frac{d\alpha }{{\alpha }^2-1}=\frac{1}{\sqrt{6}}\int (\frac{1}{\alpha -1}-\frac{1}{\alpha +1})d\alpha =\frac{1}{\sqrt{6}}ln\mid \frac{\alpha -1}{\alpha +1}\mid +C$$$$=\frac{1}{\sqrt{6}}ln\mid \frac{e^z-1}{e^z+1}\mid +Cwehavez=argsh\left(u\right)=ln(1+\sqrt{1+u^2})\Rightarrow$$$$e^z=1+\sqrt{1+u^2}=1+\sqrt{1+{\left(\sqrt{\frac{5}{6}}u\right)}^2}=1+\sqrt{1+\frac{5}{6}{u}^2}\Rightarrow$$$$I=\frac{1}{\sqrt{6}}ln\mid \frac{\sqrt{1+\frac{5}{6}{u}^2}}{2+\sqrt{1+\frac{5}{6}{u}^2}}\mid +C$$

Commented by mathmax by abdo last updated on 23/Feb/20

$$sorryz=ln(u+\sqrt{1+u^2})\Rightarrow e^z=u+\sqrt{1+u^2}=\sqrt{\frac{5}{6}}x+\sqrt{1+\frac{5}{6}{x}^2}$$$$I=\frac{1}{\sqrt{6}}ln\mid \frac{\sqrt{\frac{5}{6}}x+\sqrt{1+\frac{5}{6}{x}^2}-1}{\sqrt{\frac{5}{6}}x+\sqrt{1+\frac{5}{6}{x}^2}+1}\mid +C$$

Answered by TANMAY PANACEA last updated on 23/Feb/20

$$x=\frac{1}{t}\to dx=\frac{-dt}{t^2}$$$$\frac{2}{3}\int \frac{-dt}{t^2\times \frac{1}{t}\sqrt{\frac{5}{t^2}+6}}$$$$\frac{2}{3}\int \frac{-dt}{\sqrt{5+6t^2}}\to \frac{-2}{3\times \sqrt{6}}\int \frac{dt}{\sqrt{\frac{5}{6}+t^2}}$$$$\frac{-2}{3\sqrt{6}}ln(t+\sqrt{t^2+\frac{5}{6}})+c$$$$\frac{-2}{3\sqrt{6}}ln(\frac{1}{x}+\sqrt{\frac{1}{x^2}+\frac{5}{6}})+c$$

Commented by jagoll last updated on 23/Feb/20

$$thankyousir$$

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