Question Number 27920 by math1967 last updated on 17/Jan/18
$$\int\frac{\left({x}−\mathrm{1}\right){dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}} \\ $$
Answered by math1967 last updated on 18/Jan/18
$$\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right){dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}} \\ $$$$\int\frac{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }{dx}\:}{\frac{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}}{{x}^{\mathrm{2}} }} \\ $$$$\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}}{\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}\right)\sqrt{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}} \\ $$$${now}\:{let}\:{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}={z}^{\mathrm{2}} \therefore\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}=\mathrm{2}{zdz} \\ $$$$\int\frac{\mathrm{2}{zdz}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right).{z}}=\mathrm{2}\int\frac{{dz}}{{z}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2tan}^{−\mathrm{1}} {z}+{c} \\ $$$$\mathrm{2tan}^{−\mathrm{1}} \sqrt{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}\:+{c} \\ $$