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Question Number 67070 by mRDv143 last updated on 22/Aug/19

Commented by Prithwish sen last updated on 23/Aug/19

$$\int \frac{{\mathrm{x}}^2-1}{{(\mathrm{x}+1)}^2\sqrt{{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}}}\mathrm{dx}=\int \frac{{\mathrm{x}}^2-1}{{(\mathrm{x}+1)}^2.\mathrm{x}\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}+1}}\mathrm{dx}$$$$=\int \frac{(1-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}}{(\mathrm{x}+\frac{1}{\mathrm{x}}+2)\sqrt{\mathrm{x}+\frac{1}{\mathrm{x}}+1}}\mathrm{putting}\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{u}$$$$=\int \frac{\mathrm{du}}{(\mathrm{u}+2)\sqrt{\mathrm{u}+1}}\mathrm{putting}\mathrm{u}+1={\mathrm{t}}^2$$$$=2\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=2{\mathbf{tan}}^{-1}\mathbf{t}+\mathbf{C}=2{\mathbf{tan}}^{-1}\sqrt{\frac{{\mathbf{x}}^2+\mathbf{x}+1}{\mathbf{x}}}+\mathbf{C}$$$$\mathrm{please}\mathrm{check}.$$

Commented by mathmax by abdo last updated on 22/Aug/19

$$iseeasmallerrorifweputx+\frac{1}{x}=u\Rightarrow$$$$\int \frac{x^2-1}{{(x+1)}^2\sqrt{x^3+x^2+x}}dx=\int \frac{du}{(u+2)\sqrt{u+1}}{=}_{u+1=t^2}\int \frac{2tdt}{(t^2+1)t}$$$$=2\int \frac{dt}{1+t^2}=2arctan\left(t\right)+c=2arctan\left(\sqrt{u+1}\right)+c$$$$=2arctan\left(\sqrt{x+\frac{1}{x}+1}\right)+c$$

Commented by Prithwish sen last updated on 23/Aug/19

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}.\mathrm{It}\mathrm{was}\mathrm{a}\mathrm{typo}.\mathrm{I}$$$$\mathrm{correct}\mathrm{it}.$$

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