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dx-x-2n-1-x-n-1-

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Question Number 129646 by liberty last updated on 17/Jan/21
ϝ = ∫ (dx/(x^(2n+1) (x^n −1)))
ϝ=∫dxx2n+1(xn−1)
Answered by TheSupreme last updated on 17/Jan/21
x^n =t x=t^(1/n) x^(2n+1) =t^(2+(1/n)) dx=(1/n)t^((1/n)−1) dt ∫(((1/n)t^((1/n)−1) )/(t^(2+(1/n)) (t−1)))=∫(1/n) (1/(t^3 (t−1)))dt ((At^2 +Bt+C)/t^3 )+(D/(t−1))=(1/D) { (((t^3 ) A+D=0)),(((t^2 ) B−A=0)),(((t) −B+C=0)),(((1) −C=1)) :} F=(1/n)∫−((t^2 +t+1)/t^3 )+(1/(t−1)) dt F=(1/n)∫−(1/t)−(1/t^2 )−(1/t^3 )+(1/(t−1))dt F=(1/n)[−log(t)+(1/t)+(1/(2t^2 ))+log(t−1)+c] F=(1/n)log((x^n −1)/x^n ) +(1/x^n )+(1/(2x^(2n) ))+c
xn=tx=t1nx2n+1=t2+1ndx=1nt1n−1dt∫1nt1n−1t2+1n(t−1)=∫1n1t3(t−1)dtAt2+Bt+Ct3+Dt−1=1D{(t3)A+D=0(t2)B−A=0(t)−B+C=0(1)−C=1F=1n∫−t2+t+1t3+1t−1dtF=1n∫−1t−1t2−1t3+1t−1dtF=1n[−log(t)+1t+12t2+log(t−1)+c]F=1nlogxn−1xn+1xn+12x2n+c
Answered by liberty last updated on 17/Jan/21
ϝ = −(1/(2n)) ∫ (1/(x^n −1)) d((1/x^(2n) )) ϝ = −(1/(2n))∫ ((1/x^n )/(1−(1/x^n ))) d((1/x^(2n) )) let (1/x^n ) = r ⇒r^2 = (1/x^(2n) ) ∧ 2r dr = d((1/x^(2n) )) ϝ = −(1/(2n))∫ (r/(1−r)) (2r dr ) ϝ = −(1/n)∫ ((1−(1−r^2 ))/(1−r)) dr ϝ = (1/n) [ ln ∣1−r∣ +∫ (1+r) dr ] ϝ = (1/n) [ ln ∣1−(1/x^n )∣+(1/x^n )+(1/(2x^(2n) )) ] + C
ϝ=−12n∫1xn−1d(1x2n)ϝ=−12n∫1xn1−1xnd(1x2n)let1xn=r⇒r2=1x2n∧2rdr=d(1x2n)ϝ=−12n∫r1−r(2rdr)ϝ=−1n∫1−(1−r2)1−rdrϝ=1n[ln∣1−r∣+∫(1+r)dr]ϝ=1n[ln∣1−1xn∣+1xn+12x2n]+C
Answered by mathmax by abdo last updated on 17/Jan/21
I=∫ (dx/(x^(2n+1) (x^n −1))) changement x=t^(1/n) give I=(1/n)∫ (t^((1/n)−1) /(t^((2n+1)/n) (t−1)))dt =(1/n)∫ ((t^((1/n)−1) .t^(−2−(1/n)) )/(t−1))dt =(1/n)∫ (dt/(t^3 (t−1))) let decompose F(t)=(1/(t^3 (t−1))) F(t)=(a/t)+(b/t^2 )+(c/t^3 ) +(d/(t−1)) c=−1 ,d=1 ⇒F(t)=(a/t)+(b/t^2 )−(1/t^3 )+(1/(t−1)) lim_(t→+∞) tF(t)=0=a+1 ⇒a=−1 ⇒F(t)=−(1/t)+(b/t^2 )−(1/t^3 )+(1/(t−1)) F(−1)=(1/2)=1+b+1−(1/2)=b+2−(1/2)=b+(3/2) ⇒b=(1/2)−(3/2)=−1 ⇒ F(t)=−(1/t)−(1/t^2 )−(1/t^3 )+(1/(t−1)) ⇒∫ F(t)dt=−ln∣t∣+(1/t)+(1/(2t^2 ))+ln∣t−1∣ +C =ln∣((t−1)/t)∣+(1/t)+(1/(2t^2 )) +C =ln∣((x^n −1)/x^n )∣+(1/x^n ) +(1/(2x^(2n) )) +C
I=∫dxx2n+1(xn−1)changementx=t1ngiveI=1n∫t1n−1t2n+1n(t−1)dt=1n∫t1n−1.t−2−1nt−1dt=1n∫dtt3(t−1)letdecomposeF(t)=1t3(t−1)F(t)=at+bt2+ct3+dt−1c=−1,d=1⇒F(t)=at+bt2−1t3+1t−1limt→+∞tF(t)=0=a+1⇒a=−1⇒F(t)=−1t+bt2−1t3+1t−1F(−1)=12=1+b+1−12=b+2−12=b+32⇒b=12−32=−1⇒F(t)=−1t−1t2−1t3+1t−1⇒∫F(t)dt=−ln∣t∣+1t+12t2+ln∣t−1∣+C=ln∣t−1t∣+1t+12t2+C=ln∣xn−1xn∣+1xn+12x2n+C
Commented by mathmax by abdo last updated on 17/Jan/21
⇒I=(1/n)ln∣((x^n −1)/x^n )∣+(1/(nx^n ))+(1/(2nx^(2n) )) +C
⇒I=1nln∣xn−1xn∣+1nxn+12nx2n+C

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