Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 27920 by math1967 last updated on 17/Jan/18

∫(((x−1)dx)/((x+1)(√(x^3 +x^2 +x))))

$$\int\frac{\left({x}−\mathrm{1}\right){dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}} \\ $$

Answered by math1967 last updated on 18/Jan/18

∫(((x^2 −1)dx)/((x+1)^2 (√(x^3 +x^2 +x))))  ∫((((x^2 −1)/x^2 )dx )/(((x^2 +2x+1)(√(x^3 +x^2 +x)))/x^2 ))  ∫(((1−(1/x^2 ))dx)/((x+(1/x)+2)(√(x+(1/x)+1))))  now let x+(1/x)+1=z^2 ∴(1−(1/x^2 ))dx=2zdz  ∫((2zdz)/((z^2 +1).z))=2∫(dz/(z^2 +1))=2tan^(−1) z+c  2tan^(−1) (√(x+(1/x)+1)) +c

$$\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right){dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}} \\ $$$$\int\frac{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }{dx}\:}{\frac{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}}}{{x}^{\mathrm{2}} }} \\ $$$$\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}}{\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}\right)\sqrt{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}} \\ $$$${now}\:{let}\:{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}={z}^{\mathrm{2}} \therefore\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}=\mathrm{2}{zdz} \\ $$$$\int\frac{\mathrm{2}{zdz}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right).{z}}=\mathrm{2}\int\frac{{dz}}{{z}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2tan}^{−\mathrm{1}} {z}+{c} \\ $$$$\mathrm{2tan}^{−\mathrm{1}} \sqrt{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}\:+{c} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com