Question Number 31514 by abdo imad last updated on 09/Mar/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}. \\ $$
Answered by sma3l2996 last updated on 10/Mar/18
![by parts ∫_0 ^1 ((tan^(−1) (2x))/((1+x)^2 ))dx=[−((tan^(−1) (2x))/(1+x))]_0 ^1 +2∫(dx/((1+4x^2 )(1+x))) (1/((1+4x^2 )(1+x)))=((ax+b)/(1+4x^2 ))+(c/(1+x)) c=(1/5) ; b=(4/5) ; a=−(4/5) so ∫_0 ^1 ((tan^(−1) (2x))/((1+x)^2 ))dx=−((tan^(−1) (2))/2)+(2/5)∫_0 ^1 (((4(1−x))/(1+4x^2 ))+(1/(1+x)))dx =−((tan^(−1) (2))/2)−(8/5)∫_0 ^1 ((x−1)/(1+4x^2 ))dx+(2/5)ln(2) =(2/5)ln(2)−((tan^(−1) (2))/2)−(1/5)∫_0 ^1 (((8x)/(1+4x^2 ))−(8/(1+4x^2 )))dx =(2/5)ln(2)−((tan^(−1) (2))/2)−(1/5)[ln(1+4x^2 )]_0 ^1 +(4/5)∫_0 ^1 ((d(2x))/(1+(2x)^2 )) =(2/5)ln(2)−((tan^(−1) (2))/2)−(1/5)ln(5)+(4/5)tan^(−1) (2) =(1/5)ln((4/5))+(3/(10))tan^(−1) (2)](https://www.tinkutara.com/question/Q31556.png)
$${by}\:{parts}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}=\left[−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\mathrm{1}+{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int\frac{{dx}}{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)}=\frac{{ax}+{b}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }+\frac{{c}}{\mathrm{1}+{x}} \\ $$$${c}=\frac{\mathrm{1}}{\mathrm{5}}\:\:;\:\:{b}=\frac{\mathrm{4}}{\mathrm{5}}\:\:;\:\:{a}=−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${so} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}=−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{4}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){dx} \\ $$$$=−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{8}}{\mathrm{5}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{2}}{\mathrm{5}}{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}{ln}\left(\mathrm{2}\right)−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{8}{x}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }−\frac{\mathrm{8}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}{ln}\left(\mathrm{2}\right)−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}}\left[{ln}\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{4}}{\mathrm{5}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}{ln}\left(\mathrm{2}\right)−\frac{{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}}{ln}\left(\mathrm{5}\right)+\frac{\mathrm{4}}{\mathrm{5}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}{ln}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)+\frac{\mathrm{3}}{\mathrm{10}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$