Question Number 43417 by Raj Singh last updated on 10/Sep/18
Commented by alex041103 last updated on 10/Sep/18
$${is}\:{that}\:\sqrt[{{x}}]{{x}^{{n}} +{a}^{{n}} } \\ $$
Commented by maxmathsup by imad last updated on 10/Sep/18
$${let}\:{I}_{{n}} =\:\int\:\:\:\:\frac{{dx}}{{x}\sqrt{{x}^{{n}} \:+{a}^{{n}} }}\:\Rightarrow\:{I}_{{n}} =\:\int\:\:\:\:\:\frac{{dx}}{{x}\:{a}^{\frac{{n}}{\mathrm{2}}} \sqrt{\left(\frac{{x}}{{a}}\right)^{{n}} +\mathrm{1}}}\:\:{changement}\:\:\left(\frac{{x}}{{a}}\right)^{{n}} ={tan}^{\mathrm{2}} \theta \\ $$$${give}\:\frac{{x}}{{a}}=\left({tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow\:{dx}\:=\frac{{a}}{{n}}\:\mathrm{2}{tan}\left(\theta\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\left({tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{d}\theta \\ $$$${I}_{{n}} =\frac{\mathrm{2}{a}}{{n}}\:\int\:\:\:\:\:\:\:\frac{{tan}\theta\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\left({tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{{a}\:\left({tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{{n}}} \:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}\:{d}\theta\: \\ $$$$=\frac{\mathrm{2}}{{n}}\:\int\:\:{tan}^{−\mathrm{1}} \theta\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{d}\theta\:=\frac{\mathrm{2}}{{n}}\:\int\:\:\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta\:\frac{{sin}\theta}{{cos}\theta}}\:{d}\theta \\ $$$$=\frac{\mathrm{2}}{{n}}\:\int\:\:\:\frac{\mathrm{2}{d}\theta}{{sin}\left(\mathrm{2}\theta\right)}\:=_{\mathrm{2}\theta={u}} \:\:\frac{\mathrm{2}}{{n}}\:\int\:\:\frac{{du}}{{sinu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\frac{\mathrm{2}}{{n}}\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{{n}\:}\:\int\:\:\frac{{d}\alpha}{\alpha}\:=\frac{\mathrm{2}}{{n}}{ln}\mid\alpha\mid\:+{c}\:=\frac{\mathrm{2}}{{n}}{ln}\mid{tan}\left(\frac{{u}}{\mathrm{2}}\right)\mid\:=\frac{\mathrm{2}}{{n}}{ln}\mid{tan}\theta\mid \\ $$$$=\frac{\mathrm{2}}{{n}}{ln}\mid{tan}\left(\:{arctan}\left(\frac{{x}}{{a}}\right)^{\frac{{n}}{\mathrm{2}}} \right)\mid\:\:+{c}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
![∫(dx/(x(√(x^n +a^n )))) ∫((x^(n−1) dx)/(x^n (√(x^n +a^n )))) t^2 =x^n +a^n 2tdt=nx^(n−1) dx ∫((2tdt)/(n×(t^2 −a^n )×t)) (2/n)∫(dt/(t^2 −(a^(n/2) )^2 )) (2/n)×(1/(2a^(n/2) ))∫(((t+a^(n/2) )−(t−a^(n/2) ))/((t+a^(n/2) )(t−a^(n/2) )))dt =(1/(na^(n/2) ))[∫(dt/(t−a^(n/2) ))−∫(dt/(t+a^(n/2) ))] =(1/(na^(n/2) ))ln∣((t−a^(n/2) )/(t+a^(n/2) ))∣+c (1/(na^(n/2) ))ln∣(((√(x^n +a^n )) −a^(n/2) )/((√(x^n +a^n )) +a^(n/2) ))∣+c](Q43433.png)
$$\int\frac{{dx}}{{x}\sqrt{{x}^{{n}} +{a}^{{n}} }} \\ $$$$\int\frac{{x}^{{n}−\mathrm{1}} {dx}}{{x}^{{n}} \sqrt{{x}^{{n}} +{a}^{{n}} }} \\ $$$${t}^{\mathrm{2}} ={x}^{{n}} +{a}^{{n}} \\ $$$$\mathrm{2}{tdt}={nx}^{{n}−\mathrm{1}} {dx} \\ $$$$\int\frac{\mathrm{2}{tdt}}{{n}×\left({t}^{\mathrm{2}} −{a}^{{n}} \right)×{t}} \\ $$$$\frac{\mathrm{2}}{{n}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\left({a}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{{n}}×\frac{\mathrm{1}}{\mathrm{2}{a}^{\frac{{n}}{\mathrm{2}}} }\int\frac{\left({t}+{a}^{\frac{{n}}{\mathrm{2}}} \right)−\left({t}−{a}^{\frac{{n}}{\mathrm{2}}} \right)}{\left({t}+{a}^{\frac{{n}}{\mathrm{2}}} \right)\left({t}−{a}^{\frac{{n}}{\mathrm{2}}} \right)}{dt} \\ $$$$=\frac{\mathrm{1}}{{na}^{\frac{{n}}{\mathrm{2}}} }\left[\int\frac{{dt}}{{t}−{a}^{\frac{{n}}{\mathrm{2}}} }−\int\frac{{dt}}{{t}+{a}^{\frac{{n}}{\mathrm{2}}} }\right] \\ $$$$=\frac{\mathrm{1}}{{na}^{\frac{{n}}{\mathrm{2}}} }{ln}\mid\frac{{t}−{a}^{\frac{{n}}{\mathrm{2}}} }{{t}+{a}^{\frac{{n}}{\mathrm{2}}} }\mid+{c} \\ $$$$\frac{\mathrm{1}}{{na}^{\frac{{n}}{\mathrm{2}}} }{ln}\mid\frac{\sqrt{{x}^{{n}} +{a}^{{n}} }\:−{a}^{\frac{{n}}{\mathrm{2}}} }{\sqrt{{x}^{{n}} +{a}^{{n}} }\:+{a}^{\frac{{n}}{\mathrm{2}}} }\mid+{c} \\ $$