Question Number 120654 by peter frank last updated on 01/Nov/20
![∫_0 ^∞ (dx/([x+(√(1+x^2 )) ]^n ))](Q120654.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left[\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right]^{\mathrm{n}} } \\ $$
Answered by mindispower last updated on 01/Nov/20
![x=sh(t),n>1 ⇔∫_0 ^∞ ((ch(t)dt)/((ch(t)+sh(t))^n )) =∫_0 ^∞ ((e^t +e^(−t) )/(2e^(nt) ))dt =(1/2)[(e^((1−n)t) /(1−n))−(e^(−(1+n)t) /(1+n))]_0 ^∞ =(1/2)[(1/(n−1))+(1/(n+1))]=(n/(n^2 −1))](Q120670.png)
$${x}={sh}\left({t}\right),{n}>\mathrm{1} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\infty} \frac{{ch}\left({t}\right){dt}}{\left({ch}\left({t}\right)+{sh}\left({t}\right)\right)^{{n}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{t}} +{e}^{−{t}} }{\mathrm{2}{e}^{{nt}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{e}^{\left(\mathrm{1}−{n}\right){t}} }{\mathrm{1}−{n}}−\frac{{e}^{−\left(\mathrm{1}+{n}\right){t}} }{\mathrm{1}+{n}}\right]_{\mathrm{0}} ^{\infty} \: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right]=\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{1}} \\ $$
Commented by peter frank last updated on 01/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mindispower last updated on 03/Nov/20
$${withe}\:{pleasur} \\ $$
Answered by MJS_new last updated on 01/Nov/20
![∫(dx/((x+(√(x^2 +1)))^n ))= [t=x+(√(x^2 +1)) → dx=((√(x^2 +1))/(x+(√(x^2 +1))))dt] =(1/2)∫((t^2 +1)/t^(n+2) )dt=(1/2)∫(1/t^(n+2) )+(1/t^n )dt= =−(1/(2(n+1)t^(n+1) ))−(1/(2(n−1)t^(n−1) ))= [⇒] =−(((n+1)t^2 −(n−1))/(2(n^2 −1)t^(n+1) ))= =(((x+n(√(x^2 +1)))(−x+(√(x^2 +1)))^n )/(1−n^2 ))+C [⇒ ∫_0 ^∞ (dx/((x+(√(x^2 +1)))^n ))= =[−(1/(2(n+1)t^(n+1) ))−(1/(2(n−1)t^(n−1) ))]_1 ^∞ = =(n/(1−n^2 )) with n∈Z\{±1}](Q120671.png)
$$\int\frac{{dx}}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{{n}+\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{t}^{{n}+\mathrm{2}} }+\frac{\mathrm{1}}{{t}^{{n}} }{dt}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right){t}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right){t}^{{n}−\mathrm{1}} }=\:\left[\Rightarrow\right] \\ $$$$=−\frac{\left({n}+\mathrm{1}\right){t}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)}{\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{1}\right){t}^{{n}+\mathrm{1}} }= \\ $$$$=\frac{\left({x}+{n}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(−{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }{\mathrm{1}−{n}^{\mathrm{2}} }+{C} \\ $$$$\left[\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{n}} }=\right. \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right){t}^{{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right){t}^{{n}−\mathrm{1}} }\right]_{\mathrm{1}} ^{\infty} = \\ $$$$=\frac{{n}}{\mathrm{1}−{n}^{\mathrm{2}} }\:\mathrm{with}\:{n}\in\mathbb{Z}\backslash\left\{\pm\mathrm{1}\right\} \\ $$
Commented by peter frank last updated on 01/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$