Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 758 by 13/NaSaNa(N)056565 last updated on 08/Mar/15

∫xtan^(−1) xdx

$$\int{x}\mathrm{tan}^{−\mathrm{1}} {xdx} \\ $$$$ \\ $$

Commented by malwaan last updated on 08/Mar/15

integral by parts ∫udv=uv−∫vdu put u=tan^(−1) x ⇒du=(dx/(1+x^2 )) ; xdx=dv⇒v=(x^2 /2) and complete as prakash solution

$${integral}\:{by}\:{parts} \\ $$$$\int{udv}={uv}−\int{vdu}\: \\ $$$${put}\:{u}={tan}^{−\mathrm{1}} {x}\:\Rightarrow{du}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:;\:{xdx}={dv}\Rightarrow{v}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${and}\:{complete}\:{as}\:{prakash}\:{solution} \\ $$

Answered by prakash jain last updated on 08/Mar/15

∫tan^(−1) x ∙ x dx = (x^2 /2)tan^(−1) x −(1/2)∫(x^2 /(1+x^2 )) =(x^2 /2)tan^(−1) x −(1/2)∫ ((x^2 +1−1)/(1+x^2 )) dx =(x^2 /2)tan^(−1) x −(1/2)∫1 dx+(1/2)∫(1/(1+x^2 )) dx =(x^2 /2)tan^(−1) x −(x/2)+(1/2)tan^(−1) x+C

$$\int\mathrm{tan}^{−\mathrm{1}} {x}\:\centerdot\:{x}\:{dx}\:\: \\ $$$$=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} {x}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} {x}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} {x}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{1}\:{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} {x}\:−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} {x}+{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com