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Question Number 93610 by 675480065 last updated on 14/May/20

∫(((√(x^n +1))/x))dx

$$\int\left(\frac{\sqrt{\mathrm{x}^{\mathrm{n}} +\mathrm{1}}}{\mathrm{x}}\right)\mathrm{dx} \\ $$

Answered by 1549442205 last updated on 14/May/20

Put 1+x^n =t^n ⇒x^(n−1) dx=t^(n−1) dt.We have F(x)=∫((t^(n/2) ×t^(n−1) )/(t^n −1))dt=∫(t^((n/2)−1) +(t^((n/2)−1) /(t^n −1)))dt =_(t=u^2 ) =((2(√t^n ))/n)+∫(u^(n−2) /(u^(2n) −1))×2udu=((2(√t^n ))/n)+∫((2u^(n−1) )/(u^(2n) −1))du =((2(√t^n ))/n)+(2/n)∫(du^n /(u^(2n) −1))=_(u^n =v) ((2(√t^n ))/n)+(2/n)∫(dv/(v^2 −1)) =((2(√t^n ))/n)+(1/n)∫((1/(v−1))−(1/(v+1)))dv=((2(√(1+x^n )))/n)+ (1/n)ln∣((v−1)/(v+1))∣=((2(√(1+x^n )))/n)+(1/n)ln∣((1−(√(1+x^n )))/(1+(√(1+x^n ))))∣+C

$$\mathrm{Put}\:\mathrm{1}+\mathrm{x}^{\mathrm{n}} =\mathrm{t}^{\mathrm{n}} \Rightarrow\mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{dx}=\mathrm{t}^{\mathrm{n}−\mathrm{1}} \mathrm{dt}.\mathrm{We}\:\mathrm{have} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\int\frac{\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{2}}} ×\mathrm{t}^{\mathrm{n}−\mathrm{1}} }{\mathrm{t}^{\mathrm{n}} −\mathrm{1}}\mathrm{dt}=\int\left(\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{1}} +\frac{\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{t}^{\mathrm{n}} −\mathrm{1}}\right)\mathrm{dt} \\ $$$$=_{\mathrm{t}=\mathrm{u}^{\mathrm{2}} } =\frac{\mathrm{2}\sqrt{\mathrm{t}^{\mathrm{n}} }}{\mathrm{n}}+\int\frac{\mathrm{u}^{\mathrm{n}−\mathrm{2}} }{\mathrm{u}^{\mathrm{2n}} −\mathrm{1}}×\mathrm{2udu}=\frac{\mathrm{2}\sqrt{\mathrm{t}^{\mathrm{n}} }}{\mathrm{n}}+\int\frac{\mathrm{2u}^{\mathrm{n}−\mathrm{1}} }{\mathrm{u}^{\mathrm{2n}} −\mathrm{1}}\mathrm{du} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{t}^{\mathrm{n}} }}{\mathrm{n}}+\frac{\mathrm{2}}{\mathrm{n}}\int\frac{\mathrm{du}^{\mathrm{n}} }{\mathrm{u}^{\mathrm{2n}} −\mathrm{1}}=_{\mathrm{u}^{\mathrm{n}} =\mathrm{v}} \frac{\mathrm{2}\sqrt{\mathrm{t}^{\mathrm{n}} }}{\mathrm{n}}+\frac{\mathrm{2}}{\mathrm{n}}\int\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{t}^{\mathrm{n}} }}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}}\int\left(\frac{\mathrm{1}}{\mathrm{v}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{v}+\mathrm{1}}\right)\mathrm{dv}=\frac{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }}{\mathrm{n}}+ \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}\mathrm{ln}\mid\frac{\mathrm{v}−\mathrm{1}}{\mathrm{v}+\mathrm{1}}\mid=\frac{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}}\mathrm{ln}\mid\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }}{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }}\mid+\mathrm{C} \\ $$

Commented by 675480065 last updated on 14/May/20

if x^n =t^n −1 it means x=^n (√(t^n −1)) check ur denominator

$${if}\:{x}^{{n}} ={t}^{{n}} −\mathrm{1} \\ $$$${it}\:{means}\:{x}=^{{n}} \sqrt{{t}^{{n}} −\mathrm{1}} \\ $$$${check}\:{ur}\:{denominator} \\ $$

Commented by prakash jain last updated on 14/May/20

I think what was done was x^n +1=t^n nx^(n−1) dx=nt^(n−1) dt⇒dx=(t^(n−1) /x^(n−1) )dt ((√(x^n +1))/x)dx=(t^(n/2) /x)×(t^(n−1) /x^(n−1) )dt now you have x^n in denominator

$$\mathrm{I}\:\mathrm{think}\:\mathrm{what}\:\mathrm{was}\:\mathrm{done}\:\mathrm{was} \\ $$$${x}^{{n}} +\mathrm{1}={t}^{{n}} \\ $$$${nx}^{{n}−\mathrm{1}} {dx}={nt}^{{n}−\mathrm{1}} {dt}\Rightarrow{dx}=\frac{{t}^{{n}−\mathrm{1}} }{{x}^{{n}−\mathrm{1}} }{dt} \\ $$$$\frac{\sqrt{{x}^{{n}} +\mathrm{1}}}{{x}}{dx}=\frac{{t}^{{n}/\mathrm{2}} }{{x}}×\frac{{t}^{{n}−\mathrm{1}} }{{x}^{{n}−\mathrm{1}} }{dt} \\ $$$$\mathrm{now}\:\mathrm{you}\:\mathrm{have}\:{x}^{{n}} \:\mathrm{in}\:\mathrm{denominator} \\ $$

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