Question Number 30498 by abdo imad last updated on 22/Feb/18
$${find}\:\:{I}=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:{arcsin}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:\:. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
![by part u=arcsin(((2x)/(1+x^2 )))⇒u′=2×((1+x^2 −2x^2 )/((1+x^2 )^2 ))×(1/( (√(1−(((2x)/(1+x^2 )))^2 )))) u′=2×((1−x^2 )/((1+x^2 )^2 ))×((1+x^2 )/( (√((1+x^2 )^2 −4x^2 ))))=2×((1−x^2 )/((1+x^2 )(√(x^4 +2x^2 +1−4x^2 )))) u′=(2/(1+x^2 )) v′=1⇒v=x so I=[xarcsin(((2x)/(1+x^2 )))]_0 ^(√3) −∫_0 ^(√3) ((2x)/(1+x^2 ))dx =(√3)×(π/3)−[ln∣1+x^2 ∣]_0 ^(√3) I=((√3)/3)π−2ln(2)](https://www.tinkutara.com/question/Q30677.png)
$${by}\:{part}\:\: \\ $$$${u}={arcsin}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\Rightarrow{u}'=\mathrm{2}×\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }} \\ $$$${u}'=\mathrm{2}×\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} }}=\mathrm{2}×\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }} \\ $$$${u}'=\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$${so}\:\:{I}=\left[{xarcsin}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} −\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\sqrt{\mathrm{3}}×\frac{\pi}{\mathrm{3}}−\left[{ln}\mid\mathrm{1}+{x}^{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \\ $$$${I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\pi−\mathrm{2}{ln}\left(\mathrm{2}\right) \\ $$