Question Number 139655 by mohammad17 last updated on 30/Apr/21
Answered by qaz last updated on 30/Apr/21
![∫_0 ^1 xtan^(−1) (x)ln(x^2 +1)dx =(1/2)∫_0 ^1 tan^(−1) (x)ln(x^2 +1)d(x^2 +1) =(1/2)∫_0 ^1 tan^(−1) (x)d[(x^2 +1)ln(x^2 +1)−(x^2 +1)] =(1/2){(π/4)(2ln2−2)−∫_0 ^1 (([(x^2 +1)ln(x^2 +1)−(x^2 +1)])/(x^2 +1))dx} =(1/2){(π/2)ln(2/e)−∫_0 ^1 [ln(x^2 +1)−1]dx} =(π/4)ln(2/e)+(1/2)−(1/2){ln2−∫_0 ^1 ((2x^2 )/(x^2 +1))dc} =(π/4)ln(2/e)+(1/2)−(1/2)ln2+1−(π/4) =(π/4)ln(2/e^2 )+(3/2)−(1/2)ln2](https://www.tinkutara.com/question/Q139691.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\mathrm{tan}^{−\mathrm{1}} \left({x}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right){d}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left({x}\right){d}\left[\left({x}^{\mathrm{2}} +\mathrm{1}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{4}}\left(\mathrm{2}{ln}\mathrm{2}−\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left[\left({x}^{\mathrm{2}} +\mathrm{1}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{2}}{ln}\frac{\mathrm{2}}{{e}}−\int_{\mathrm{0}} ^{\mathrm{1}} \left[{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{1}\right]{dx}\right\} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\frac{\mathrm{2}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mathrm{2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dc}\right\} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\frac{\mathrm{2}}{{e}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\mathrm{1}−\frac{\pi}{\mathrm{4}} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\frac{\mathrm{2}}{{e}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\ $$