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Question Number 86956 by abdomathmax last updated on 01/Apr/20

find ∫(1−(1/x^2 ))arctan(2x)dx

$${find}\:\int\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right){dx} \\ $$

Commented by Ar Brandon last updated on 01/Apr/20

Let u=arctan(2x) ⇒ du=2∙(1/(1+(2x)^2 ))dx dv=(1−(1/x^2 ))dx ⇒ v=(x+(1/x)) I=(x+(1/x^2 ))arctan(2x)−2∫(x+(1/x))∙((1/(1+4x^2 )))dx =(x+(1/x^2 ))arctan(2x)−2∫((x^2 +1)/x)∙(1/(1+4x^2 ))dx =(x+(1/x^2 ))arctan(2x)−2∫((x^2 +1)/(x(1+4x^2 )))dx ((x^2 +1)/(x(1+4x^2 )))=(A/x)+((Bx+C)/(4x^2 +1)) =((A(4x^2 +1)+(Bx+C)x)/(x(4x^2 +1)))=(((4A+B)x^2 +Cx+A)/(x(4x^2 +1))) 4A+B=1 C=0 A=1 ⇒ B=−3 ∫(x^2 /(x(1+4x^2 )))dx=∫((1/x)−((3x)/(1+4x^2 )))=∫(1/x)dx−(3/8)∫((8x)/(1+4x^2 ))dx =lnx−(3/8)ln(1+4x^2 )+Constant ∫(1−(1/x^2 ))arctan(2x)dx=(x+(1/x))arctan(2x)−2lnx+(3/4)ln(1+4x^2 )+constant

$${Let}\:{u}={arctan}\left(\mathrm{2}{x}\right)\:\Rightarrow\:{du}=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:{dv}=\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}\:\:\Rightarrow\:\:{v}=\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$ \\ $$$${I}=\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right)−\mathrm{2}\int\left({x}+\frac{\mathrm{1}}{{x}}\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right){dx} \\ $$$$\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right)−\mathrm{2}\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\centerdot\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right)−\mathrm{2}\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}{dx} \\ $$$$ \\ $$$$ \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}=\frac{{A}}{{x}}+\frac{{Bx}+{C}}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{A}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)+\left({Bx}+{C}\right){x}}{{x}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\left(\mathrm{4}{A}+{B}\right){x}^{\mathrm{2}} +{Cx}+{A}}{{x}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{4}{A}+{B}=\mathrm{1} \\ $$$${C}=\mathrm{0} \\ $$$${A}=\mathrm{1}\:\Rightarrow\:\:{B}=−\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$\int\frac{{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)}{dx}=\int\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{3}{x}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)=\int\frac{\mathrm{1}}{{x}}{dx}−\frac{\mathrm{3}}{\mathrm{8}}\int\frac{\mathrm{8}{x}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={lnx}−\frac{\mathrm{3}}{\mathrm{8}}{ln}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)+{Constant} \\ $$$$ \\ $$$$\int\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left(\mathrm{2}{x}\right){dx}=\left({x}+\frac{\mathrm{1}}{{x}}\right){arctan}\left(\mathrm{2}{x}\right)−\mathrm{2}{lnx}+\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)+{constant} \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 01/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Ar Brandon last updated on 01/Apr/20

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