dx-x-n-x-the-problems-i-have-posted-are-tricky-

Question Number 47743 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

$$\int\frac{{dx}}{{x}^{{n}} −{x}}\:\: \\ $$$${the}\:{problems}\:{i}\:{have}\:{posted}\:{are}\:{tricky}… \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

∫(dx/(x(x^(n−1) −1))) ∫((x^(n−2) dx)/(x^(n−1) (x^(n−1) −1))) t=x^(n−1) dt=(n−1)x^(n−2) dx ∫(dt/((n−1)(t)(t−1))) (1/(n−1))∫((t−(t−1))/(t(t−1)))dt (1/(n−1))[∫(dt/(t−1))−∫(dt/t)] (1/(n−1))[ln(((t−1)/t))]+c (1/(n−1))[ln(((x^(n−1) −1)/x^(n−1) ))]+c

$$\int\frac{{dx}}{{x}\left({x}^{{n}−\mathrm{1}} −\mathrm{1}\right)} \\ $$$$\int\frac{{x}^{{n}−\mathrm{2}} \:{dx}}{{x}^{{n}−\mathrm{1}} \left({x}^{{n}−\mathrm{1}} −\mathrm{1}\right)} \\ $$$${t}={x}^{{n}−\mathrm{1}} \:\:\:{dt}=\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {dx} \\ $$$$\int\frac{{dt}}{\left({n}−\mathrm{1}\right)\left({t}\right)\left({t}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{{n}−\mathrm{1}}\int\frac{{t}−\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{{n}−\mathrm{1}}\left[\int\frac{{dt}}{{t}−\mathrm{1}}−\int\frac{{dt}}{{t}}\right] \\ $$$$\frac{\mathrm{1}}{{n}−\mathrm{1}}\left[{ln}\left(\frac{{t}−\mathrm{1}}{{t}}\right)\right]+{c} \\ $$$$\frac{\mathrm{1}}{{n}−\mathrm{1}}\left[{ln}\left(\frac{{x}^{{n}−\mathrm{1}} −\mathrm{1}}{{x}^{{n}−\mathrm{1}} }\right)\right]+{c} \\ $$$$ \\ $$

Answered by ajfour last updated on 14/Nov/18

let t=(1/x) ⇒ dt = −(dx/x^2 ) dx = −(dt/t^2 ) I = ∫(((−(dt/t^2 )))/((1/t^n )−(1/t))) = −∫((t^(n−2) dt)/(1−t^(n−1) )) = (1/(n−1))∫ ((−(n−1)t^(n−2) dt)/(1−t^(n−1) )) I = (1/(n−1))ln ∣1−(1/t^(n−1) )∣+c .

$${let}\:\:\:{t}=\frac{\mathrm{1}}{{x}}\:\:\:\Rightarrow\:\:{dt}\:=\:−\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:{dx}\:=\:−\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$\:\:{I}\:=\:\int\frac{\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)}{\frac{\mathrm{1}}{{t}^{{n}} }−\frac{\mathrm{1}}{{t}}}\:=\:−\int\frac{{t}^{{n}−\mathrm{2}} {dt}}{\mathrm{1}−{t}^{{n}−\mathrm{1}} } \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{{n}−\mathrm{1}}\int\:\frac{−\left({n}−\mathrm{1}\right){t}^{{n}−\mathrm{2}} {dt}}{\mathrm{1}−{t}^{{n}−\mathrm{1}} } \\ $$$$\:\:\:\:{I}\:=\:\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{ln}\:\mid\mathrm{1}−\frac{\mathrm{1}}{{t}^{{n}−\mathrm{1}} }\mid+{c}\:. \\ $$$$ \\ $$

Commented by rahul 19 last updated on 14/Nov/18

Nice �� Is there any particular reason why u substitute t=1/x?

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

$${excellent}… \\ $$

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