Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 171664 by LEKOUMA last updated on 19/Jun/22

Solve  ∫(x^2 /(1+x^2 ))tan^(−1) xdx

$${Solve} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} {xdx} \\ $$

Commented by infinityaction last updated on 19/Jun/22

put   tan^(−1) x = y  ⇒ x^2  = tan^2  y  (1/(1+x^2 ))dx = dy  ∫ytan^2  y dy  by parts

$${put}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\:=\:{y}\:\:\Rightarrow\:{x}^{\mathrm{2}} \:=\:\mathrm{tan}^{\mathrm{2}} \:{y} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:{dy} \\ $$$$\int{y}\mathrm{tan}^{\mathrm{2}} \:{y}\:{dy} \\ $$$${by}\:{parts} \\ $$

Answered by mindispower last updated on 19/Jun/22

x^2 =1+x^2 −1  =∫tg^− (x)dx−∫tg^− (x)d(tg^− (x))dx  =xtg^− (x)−∫(x/(1+x^2 ))dx−(1/2)(tg^− (x))^2   =xtg^− (x)−(1/2)(tg−(x))^2 −(1/2)ln(1+x^2 )+c

$${x}^{\mathrm{2}} =\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\int{tg}^{−} \left({x}\right){dx}−\int{tg}^{−} \left({x}\right){d}\left({tg}^{−} \left({x}\right)\right){dx} \\ $$$$={xtg}^{−} \left({x}\right)−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\left({tg}^{−} \left({x}\right)\right)^{\mathrm{2}} \\ $$$$={xtg}^{−} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({tg}−\left({x}\right)\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{c} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com