Question Number 30518 by abdo imad last updated on 22/Feb/18
$${let}\:{a}>\mathrm{0}\:{find}\:\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}+{a}\right)\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }}\:. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{a}^{\mathrm{2}} \left(\frac{{x}}{{a}}+\mathrm{1}\right)\sqrt{\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\mathrm{1}}} \\ $$$${let}\:\:\frac{{x}}{{a}}={sinh}\left({t}\right)\Rightarrow{dx}={a}×{cosh}\left({t}\right){dt}={a}\sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} {t}}{dt} \\ $$$${dt}=\frac{{dx}}{{a}\sqrt{\mathrm{1}+\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} }} \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({sinh}\left({t}\right)+\mathrm{1}\right)}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{t}} }{{e}^{\mathrm{2}{t}} +{e}^{{t}} −\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{t}} }{\left(\frac{\mathrm{2}{e}^{{t}} +\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}}{dt}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{t}} }{\frac{\mathrm{5}}{\mathrm{4}}\left(\left(\frac{\mathrm{2}{e}^{{t}} +\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} −\mathrm{1}\right)}{dt} \\ $$$${u}=\frac{\mathrm{2}{e}^{{t}} +\mathrm{1}}{\:\sqrt{\mathrm{5}}}\Rightarrow{du}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}{e}^{{t}} {dt} \\ $$$${e}^{{t}} {dt}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{du} \\ $$$${f}\left({a}\right)=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}{a}}\int_{\mathrm{1}/\sqrt{\mathrm{5}}} ^{\infty} \frac{{du}}{{u}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}{a}}\int_{\mathrm{1}/\sqrt{\mathrm{5}}} ^{\infty} \frac{{du}}{\left({u}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)} \\ $$$${f}\left({a}\right)=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}{a}}\left[{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\right]_{\mathrm{1}/\sqrt{\mathrm{5}}} ^{\infty} =−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}{a}}{ln}\mid\frac{\mathrm{1}/\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{1}/\sqrt{\mathrm{5}}+\mathrm{1}}\mid \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}{a}}{ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{1}−\sqrt{\mathrm{5}}}\mid=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}{a}}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$ \\ $$