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 51421 by Tawa1 last updated on 26/Dec/18

∫ ((tan^(−1) x)/x^2 ) dx

$$\int\:\:\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{dx} \\ $$

Commented by Tawa1 last updated on 27/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 26/Dec/18

let determine f(t) =∫ ((arctan(tx))/x^2 )dx with t>0 by parts f(t) =−(1/x) arctan(tx) −∫ −(1/x) (t/(1+t^2 x^2 ))dx =−((arctan(tx))/x) + t ∫ (dx/(x(1+t^2 x^2 ))) but ∫ (dx/(x(1+t^2 x^2 ))) =_(tx=u) ∫ (du/(t(u/t)(1+u^2 ))) = ∫ (du/(u(1+u^2 ))) =∫ ((1/u) −(u/(1+u^2 ))) =ln∣u∣ −(1/2)ln(1+u^2 ) = ln(((∣u∣)/(√(1+u^2 )))) +c =ln(((∣tx∣)/(√(1+t^2 x^2 )))) ⇒f(t) =−((arctan(tx))/x) +t ln(((∣tx∣)/(√(1+t^2 x^2 ))))+c and ∫ ((arctan(x))/x^2 )dx =f(1) =−((arctanx)/x) +ln(((∣x∣)/(√(1+x^2 ))))+C .

$${let}\:{determine}\:{f}\left({t}\right)\:=\int\:\:\frac{{arctan}\left({tx}\right)}{{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}>\mathrm{0}\:\:{by}\:{parts} \\ $$$${f}\left({t}\right)\:=−\frac{\mathrm{1}}{{x}}\:{arctan}\left({tx}\right)\:\:−\int\:\:−\frac{\mathrm{1}}{{x}}\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{{arctan}\left({tx}\right)}{{x}}\:+\:{t}\:\int\:\:\:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}\:{but} \\ $$$$\int\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}\:=_{{tx}={u}} \:\:\:\:\:\int\:\:\frac{{du}}{{t}\frac{{u}}{{t}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\:\int\:\:\:\frac{{du}}{{u}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\int\:\left(\frac{\mathrm{1}}{{u}}\:−\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right)\:={ln}\mid{u}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\:=\:{ln}\left(\frac{\mid{u}\mid}{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\right)\:+{c} \\ $$$$={ln}\left(\frac{\mid{tx}\mid}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} \:{x}^{\mathrm{2}} }}\right)\:\Rightarrow{f}\left({t}\right)\:=−\frac{{arctan}\left({tx}\right)}{{x}}\:+{t}\:{ln}\left(\frac{\mid{tx}\mid}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} }}\right)+{c}\:\:{and} \\ $$$$\int\:\:\frac{{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{1}\right)\:=−\frac{{arctanx}}{{x}}\:+{ln}\left(\frac{\mid{x}\mid}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)+{C}\:. \\ $$

Commented by Abdo msup. last updated on 30/Dec/18

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

Answered by ajfour last updated on 26/Dec/18

I =−((tan^(−1) x)/x)+∫(dx/(x(1+x^2 ))) Now (1/(x(1+x^2 ))) = (A/x)+((Bx+C)/(1+x^2 )) ⇒ 1= A+Ax^2 +Bx^2 +Cx ⇒ A=1, C = 0, B = −1 I =−((tan^(−1) x)/x)+∫ (dx/x)−∫(x/(1+x^2 )) dx I = −((tan^(−1) x)/x)+ln ∣x∣−(1/2)ln ∣1+x^2 ∣+c .

$${I}\:=−\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}+\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${Now}\:\:\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\frac{{A}}{{x}}+\frac{{Bx}+{C}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\mathrm{1}=\:{A}+{Ax}^{\mathrm{2}} +{Bx}^{\mathrm{2}} +{Cx} \\ $$$$\Rightarrow\:\:{A}=\mathrm{1},\:\:{C}\:=\:\mathrm{0},\:{B}\:=\:−\mathrm{1} \\ $$$${I}\:=−\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}+\int\:\frac{{dx}}{{x}}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:{I}\:=\:−\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}+\mathrm{ln}\:\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{1}+{x}^{\mathrm{2}} \mid+{c}\:. \\ $$

Commented by Tawa1 last updated on 26/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Smail last updated on 26/Dec/18

by parts u=tan^(−1) x⇒u′=(1/(1+x^2 )) v′=(1/x^2 )⇒v=((−1)/x) A=∫((tan^(−1) x)/x^2 )dx=((−tan^(−1) x)/x)+∫(dx/(x(1+x^2 )))+c_1 (1/(x(1+x^2 )))=(a/x)+((bx+c)/(x^2 +1)) with a=1 , b=−1 , c=0 A=((−tan^(−1) x)/x)+∫((1/x)−(x/(x^2 +1)))dx+c_1 =−((tan^(−1) x)/x)+ln∣x∣−(1/2)ln(x^2 +1)+C ∫((tan^(−1) x)/x^2 )dx=−((tan^(−1) x)/x)+ln∣(x/(√(x^2 +1)))∣+C

$${by}\:{parts} \\ $$$${u}={tan}^{−\mathrm{1}} {x}\Rightarrow{u}'=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${v}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\Rightarrow{v}=\frac{−\mathrm{1}}{{x}} \\ $$$${A}=\int\frac{{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} }{dx}=\frac{−{tan}^{−\mathrm{1}} {x}}{{x}}+\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}+{c}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}=\frac{{a}}{{x}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:{with}\:\:\:\:{a}=\mathrm{1}\:\:,\:\:{b}=−\mathrm{1}\:,\:{c}=\mathrm{0} \\ $$$${A}=\frac{−{tan}^{−\mathrm{1}} {x}}{{x}}+\int\left(\frac{\mathrm{1}}{{x}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}+{c}_{\mathrm{1}} \\ $$$$=−\frac{{tan}^{−\mathrm{1}} {x}}{{x}}+{ln}\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{C} \\ $$$$\int\frac{{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} }{dx}=−\frac{{tan}^{−\mathrm{1}} {x}}{{x}}+{ln}\mid\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\mid+{C} \\ $$

Commented by Tawa1 last updated on 26/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com