Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 66096 by sitangshu17 last updated on 09/Aug/19

$$\int \frac{\sqrt{(1+{\mathrm{x}}^2)}}{{\mathrm{x}}^2}\mathrm{dx}=?$$

Commented by mathmax by abdo last updated on 09/Aug/19

$$letI=\int \frac{\sqrt{1+x^2}}{x^2}dxbypartsI=-\frac{1}{x}\sqrt{1+x^2}-\int (-\frac{1}{x})\frac{2x}{2\sqrt{1+x^2}}dx$$$$=-\frac{\sqrt{1+x^2}}{x}+\int \frac{dx}{\sqrt{1+x^2}}=-\frac{\sqrt{1+x^2}}{x}+ln(x+\sqrt{1+x^2})+C$$

Commented by Prithwish sen last updated on 09/Aug/19

$$\mathrm{put}1+{\mathrm{x}}^2={\mathrm{x}}^2{\mathrm{z}}^2$$$$\frac{\mathrm{dx}}{\mathrm{x}}=\frac{\mathrm{zdz}}{1-{\mathrm{z}}^2}$$$$\int \frac{\sqrt{1+{\mathrm{x}}^2}}{{\mathrm{x}}^2}\mathrm{dx}=\int \frac{\mathrm{xz}}{{\mathrm{x}}^2}\mathrm{dx}=\int \frac{\mathrm{zdx}}{\mathrm{x}}=\int \frac{{\mathrm{z}}^2}{1-{\mathrm{z}}^2}\mathrm{dz}$$$$=\int [\frac{1}{1-{\mathrm{z}}^2}-1]\mathrm{dz}=\frac{1}{2}\ln \mid \frac{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}-\sqrt{1+{\mathrm{x}}^2}}\mid -\frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}}+\mathrm{C}$$$$\mathrm{by}\mathrm{putting}\mathrm{z}=\frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}}\mathrm{please}\mathrm{check}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com