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Question Number 57433 by naka3546 last updated on 04/Apr/19

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

$$\int \frac{x^4+1}{x^2\sqrt{x^2(x^2-\frac{1}{x^2})}}dx$$$$\int \frac{x+\frac{1}{x^3}}{\sqrt{x^2-\frac{1}{x^2}}}dx$$$$t^2=x^2-\frac{1}{x^2}$$$$2tdt=(2x+\frac{2}{x^3})dx$$$$tdt=(x+\frac{1}{x^3})dx$$$$\int \frac{tdt}{t}$$$$=t+c$$$$=\sqrt{x^2-\frac{1}{x^2}}+c$$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

$$plscheck...trickyproblem...$$

Commented by MJS last updated on 04/Apr/19

$$\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{correct}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19

$$thankyousir...$$

Commented by malwaan last updated on 05/Apr/19

$$\mathit{a}\mathit{n}\mathit{d}\sqrt{{\mathit{x}}^2-\frac{1}{{\mathit{x}}^2}}=\frac{\sqrt{{\mathit{x}}^4-1}}{\mathit{x}}$$$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{s}\mathit{o}\mathit{m}\mathit{u}\mathit{c}\mathit{h}$$

Answered by MJS last updated on 04/Apr/19

$$\mathrm{possible}\mathrm{hat}\mathrm{trick}:$$$$\int \frac{x^4+1}{x^2\sqrt{x^4-1}}dx=\int \frac{2x^2}{\sqrt{x^4-1}}dx-\int \frac{\sqrt{x^4-1}}{x^2}dx$$$$\mathrm{now}\mathrm{write}\int \frac{\sqrt{x^4-1}}{x^2}dx\mathrm{by}\mathrm{parts}\mathrm{without}$$$$\mathrm{solving}:$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$$$u^{\prime }=\frac{1}{x^2}\Rightarrow u=-\frac{1}{x}$$$$v=\sqrt{x^4-1}\Rightarrow v^{\prime }=\frac{2x^3}{\sqrt{x^4-1}}$$$$\mathrm{so}\mathrm{we}\mathrm{have}:$$$$\int \frac{x^4+1}{x^2\sqrt{x^4-1}}dx=\int \frac{2x^2}{\sqrt{x^4-1}}dx-(-\frac{\sqrt{x^4-1}}{x}-\int -\frac{2x^2}{\sqrt{x^4-1}}dx)=$$$$=\frac{\sqrt{x^4-1}}{x}+C$$

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