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Question Number 36801 by rahul 19 last updated on 05/Jun/18

$$\int \frac{1+x^4}{{(1-x^4)}^{\frac{3}{2}}}dx=A$$$$\int \mathrm{A}=\mathrm{B}$$$$\mathrm{Find}\mathrm{B}?$$$$\mathrm{Assume}\mathrm{integration}\mathrm{of}\mathrm{constant}=0.$$

Answered by MJS last updated on 05/Jun/18

$$A=\frac{x}{\sqrt{1-x^4}}$$$$B=\frac{1}{2}\arcsin x^2$$$$$$$$A:$$$$\int \frac{1+x^4}{{(1-x^4)}^{\frac{3}{2}}}dx$$$$\mathrm{looks}\mathrm{like}{\mathrm{it}}^{\prime }\mathrm{s}\frac{p\left(x\right)}{\sqrt{1-x^4}}\mathrm{with}p\left(x\right)\mathrm{is}\mathrm{a}\mathrm{polynome}$$$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{try}$$$$\frac{d}{dx}\left(\frac{p\left(x\right)}{\sqrt{1-x^4}}\right)=\frac{d}{dx}\left(p\left(x\right){(1-x^4)}^{-\frac{1}{2}}\right)=$$$$=p^{\prime }\left(x\right){(1-x^4)}^{-\frac{1}{2}}+p\left(x\right)\left(2x^3{(1-x^4)}^{-\frac{3}{2}}\right)=$$$$=\frac{p^{\prime }\left(x\right)(1-x^4)+2p\left(x\right)x^3}{{(1-x^4)}^{\frac{3}{2}}}$$$$\Rightarrow p^{\prime }\left(x\right)(1-x^4)+2p\left(x\right)x^3=1+x^4$$$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{try}p\left(x\right)=ax+b;p^{\prime }\left(x\right)=a$$$$a+2{bx}^3+{ax}^4=1+x^4$$$$a=1;b=0$$$$p\left(x\right)=x$$$$\int \frac{1+x^4}{{(1-x^4)}^{\frac{3}{2}}}dx=\frac{x}{\sqrt{1-x^4}}$$$$$$$$B:$$$$\int \frac{x}{\sqrt{1-x^4}}dx=$$$$[t=x^2\to dx=\frac{dt}{2x}]$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{1-t^2}}=\frac{1}{2}\arcsin t=$$$$=\frac{1}{2}\arcsin x^2$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

$$$$$$\int \frac{1+x^4}{(1-x^4)\sqrt{1-x^4}}dx$$$$x^2=\frac{1}{t}x=t^{\frac{-1}{2}}dx=\frac{-1}{2}\times \frac{1}{t^{\frac{3}{2}}}dt$$$$=\frac{-1}{2}\int \frac{dt}{t^{\frac{3}{2}}}\times \frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}\times \frac{1}{\sqrt{1-\frac{1}{t^2}}}$$$$=\frac{-1}{2}\int \frac{dt}{t^{\frac{3}{2}}}\times \frac{t^2+1}{t^2-1}\times \frac{t}{\sqrt{t^2-1}}$$$$=\frac{-1}{2}\int \frac{t^2+1}{t^2-1}\times \frac{1}{\sqrt{t}}\times \frac{1}{\sqrt{t}(\sqrt{t-\frac{1}{t}}}dt$$$$=\frac{-1}{2}\int \frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}\times \frac{1}{t}\times \frac{dt}{\sqrt{t-\frac{1}{t}}}$$$$\frac{-1}{2}\int \frac{1+\frac{1}{t^2}}{t-\frac{1}{t}}\times \frac{dt}{\sqrt{t-\frac{1}{t}}}$$$$=\frac{-1}{2}\int \frac{dk}{k^{\frac{3}{2}}}when(k=t-\frac{1}{t})$$$$=\frac{-1}{2}\times \frac{k^{\frac{-1}{2}}}{\frac{-1}{2}}+c$$$$=\frac{1}{\sqrt{k}}+c$$$$=\frac{1}{\sqrt{t-\frac{1}{t}}}+c$$$$=\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}+c$$$$$$$$soA=\frac{1}{\sqrt{\frac{1}{x^2}-x^2}}$$$$\int Adx=B$$$$\int \frac{x}{\sqrt{1-x^4}}dx$$$$y=x^2$$$$dy=2xdx$$$$\int \frac{dy}{2\times \sqrt{1-y^2}}$$$$=\frac{1}{2}{sin}^{-1}\left(y\right)$$$$=\frac{1}{2}{sin}^{-1}\left(x^2\right)isB$$$$$$

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