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0-1-3-x-2n-ln-1-x-dx-

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Question Number 128633 by Lordose last updated on 09/Jan/21
Ω = ∫_0 ^( (1/3)) x^(2n) ln(1−x)dx
Ω=∫013x2nln(1−x)dx
Answered by mathmax by abdo last updated on 09/Jan/21
let try another way x=(t/3) ⇒ Ω=∫_0 ^1 ((t/3))^(2n) ln(1−(t/3))(dt/3) =(1/3^(2n+1) )∫_0 ^1 t^(2n) ln(((3−t)/3))dt =(1/3^(2n+1) ) ∫_0 ^1 t^(2n) ln(3−t)dt−(1/3^(2n+1) )ln(3)[(t^(2n+1) /(2n+1))]_0 ^1 =(1/3^(2n+1) )∫_0 ^1 t^(2n) ln(3−t)dt −((ln3)/((2n+1)3^(2n+1) )) we have by parts ∫_0 ^1 t^(2n) ln(3−t)dt =[(t^(2n+1) /(2n+1))ln(3−t)]_0 ^1 −∫_0 ^1 (t^(2n+1) /(2n+1))×((−1)/(3−t))dt =(1/(2n+1))ln(2)+(1/(2n+1))∫_0 ^1 (t^(2n+1) /(3−t))dt and ∫_0 ^1 (t^(2n+1) /(3−t)) dt =_(3−t=u) −∫_2 ^3 (((3−u)^(2n+1) )/u)(−du) =−∫_2 ^3 (((u−3)^(2n+1) )/u) du =−∫_2 ^3 ((Σ_(k=0) ^(2n+1) C_(2n+1) ^k u^k (−3)^(2n+1−k) )/u)du =3^(2n+1) Σ_(k=1) ^(2n+1) (−3)^(−k) C_(2n+1) ^k ∫_2 ^3 u^(k−1) du −3^(2n+1) ln((3/2)) =3^(2n+1) Σ_(k=0) ^(2n+1) (−3)^(−k) C_(2n+1) ^k (1/k){ 3^(k−1) −2^(k−1) }−3^(2n+1) ln((3/2))...
lettryanotherwayx=t3⇒Ω=∫01(t3)2nln(1−t3)dt3=132n+1∫01t2nln(3−t3)dt=132n+1∫01t2nln(3−t)dt−132n+1ln(3)[t2n+12n+1]01=132n+1∫01t2nln(3−t)dt−ln3(2n+1)32n+1wehavebyparts∫01t2nln(3−t)dt=[t2n+12n+1ln(3−t)]01−∫01t2n+12n+1×−13−tdt=12n+1ln(2)+12n+1∫01t2n+13−tdtand∫01t2n+13−tdt=3−t=u−∫23(3−u)2n+1u(−du)=−∫23(u−3)2n+1udu=−∫23∑k=02n+1C2n+1kuk(−3)2n+1−kudu=32n+1∑k=12n+1(−3)−kC2n+1k∫23uk−1du−32n+1ln(32)=32n+1∑k=02n+1(−3)−kC2n+1k1k{3k−1−2k−1}−32n+1ln(32)…
Answered by mathmax by abdo last updated on 09/Jan/21
Ω=∫_0 ^(1/3) x^(2n) ln(1−x)dx by parts u^′ =x^(2n) [and v=ln(1−x) Ω=[(x^(2n+1) /(2n+1))ln(1−x)]_0 ^(1/3) −∫_0 ^(1/3) (x^(2n+1) /(2n+1))×((−1)/(1−x))dx =(1/(2n+1(3)^(2n+1) ))ln((2/3))+(1/(2n+1))∫_0 ^(1/3) (x^(2n+1) /(1−x))dx but ∫_0 ^(1/3) (x^(2n+1) /(1−x))dx =∫_0 ^(1/3) ((x^(2n+1) −1+1)/(1−x))dx =∫_0 ^(1/3) (((x−1)(1+x+x^2 +...+x^(2n) ))/(1−x))dx +[−ln(1−x)]_0 ^(1/3) =−∫_0 ^(1/3) (Σ_(k=0) ^(2n) x^k )dx−ln((2/3)) =−Σ_(k=0) ^(2n) [(x^(k+1) /(k+1))]_0 ^(1/3) −ln2 +ln3 =−Σ_(k=0) ^(2n) (1/((k+1)3^(k+1) )) +ln3−ln2 rsst to find the value of this serie...
Ω=∫013x2nln(1−x)dxbypartsu′=x2n[andv=ln(1−x)Ω=[x2n+12n+1ln(1−x)]013−∫013x2n+12n+1×−11−xdx=12n+1(3)2n+1ln(23)+12n+1∫013x2n+11−xdxbut∫013x2n+11−xdx=∫013x2n+1−1+11−xdx=∫013(x−1)(1+x+x2+…+x2n)1−xdx+[−ln(1−x)]013=−∫013(∑k=02nxk)dx−ln(23)=−∑k=02n[xk+1k+1]013−ln2+ln3=−∑k=02n1(k+1)3k+1+ln3−ln2rssttofindthevalueofthisserie…

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