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Question Number 38746 by MJS last updated on 29/Jun/18

$$\mathrm{this}\mathrm{is}\mathrm{still}\mathrm{waiting}\mathrm{to}\mathrm{be}\mathrm{solved}...$$$$\int \frac{\sqrt{(t-1)t(t+1)}}{3t^2-4}dt=?$$

Commented by behi83417@gmail.com last updated on 29/Jun/18

$$thiscannotbedifinedintermsof$$$$primaryfunctions.dontspendtimeon$$$$it.$$

Commented by prof Abdo imad last updated on 29/Jun/18

$$I=\int \frac{\sqrt{t(t^2-1)}}{3t^2-4}dtchangementt=ch\left(x\right)give$$$$I=\int \frac{\sqrt{ch\left(x\right)}sh\left(x\right)}{3{ch}^{2}x-4}sh\left(x\right)dx$$$$=\int \frac{{sh}^2\left(x\right)\sqrt{ch\left(x\right)}}{3{ch}^{2}x-4}dx\sqrt{chx}=u\Rightarrow chx=u^2\Rightarrow x=argch\left(u^2\right)$$$$I=\int \frac{(u^4-1)u}{3u^4-4}\frac{2udu}{\sqrt{u^4-1}}$$$$=\int \frac{2u^2\sqrt{u^4-1}}{3u^4-4}du=\int \frac{2u^2\sqrt{u^4-1}}{4u^4-4-u^4}du$$$$=\int \frac{2u^2\sqrt{u^4-1}}{4{\left(\sqrt{u^4-1}\right)}^2-u^4}duchang.\sqrt{u^4-1}=x\Rightarrow u^4-1=x^2\Rightarrow u={(1+x^2)}^{\frac{1}{4}}$$$$I=\int \frac{x\sqrt{x^2+1}}{4x^2-1-x^2}\frac{x}{2}{(1+x^2)}^{-\frac{3}{4}}dx$$$$=\int \frac{x^2{(x^2+1)}^{-\frac{1}{4}}}{3x^2-1}dxchang.x=tan\theta give$$$$I=\int \frac{{tan}^2\theta {\left({cos}^{-2}\theta \right)}^{-\frac{1}{4}}}{3{tan}^2\theta -1}(1+{tan}^2\theta )d\theta$$$$....becontinued...$$

Commented by MJS last updated on 29/Jun/18

$$\mathrm{Sir}\mathrm{Behi},\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{same}.\mathrm{but}\mathrm{can}\mathrm{we}\mathrm{be}\mathrm{sure}?$$

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