Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 125911 by Lordose last updated on 15/Dec/20

∫_0 ^( ∞) ((xln(1+x))/(1+x^4 ))dx

0xln(1+x)1+x4dx

Answered by mathmax by abdo last updated on 15/Dec/20

A =∫_0 ^∞  ((xln(1+x))/(1+x^4 ))dx =∫_0 ^1  ((xln(1+x))/(1+x^4 ))dx+∫_1 ^∞  ((xln(x+1))/(1+x^4 ))dx  ∫_1 ^∞  ((xln(x+1))/(1+x^4 ))dx =_(x=(1/t))   −∫_0 ^1  ((ln(1+(1/t)))/(t(1+(1/t^4 ))))(−(dt/t^2 ))  =∫_0 ^1  ((ln(t+1)−ln(t)/(t^3  +(1/t)))dt =∫_0 ^1  ((tln(t+1)−tlnt)/(1+t^4 ))dt  ⇒A =2∫_0 ^1  ((xln(x+1))/(1+x^4 ))dx−∫_0 ^1  ((xlnx)/(1+x^4 ))dx  we have  ∫_0 ^1  ((xlnx)/(1+x^4 ))dx =∫_0 ^1 xln(x)Σ_(n=0) ^∞ (−1)^n  x^(4n) dx  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^1  x^(4n+1) ln(x)dx =Σ_(n=0) ^∞  (−1)^n  u_n   u_n =∫_0 ^1  x^(4n+1) ln(x)dx =[(1/(4n+2))x^(4n+2) ln(x)]_0 ^1 −∫_0 ^1 (x^(4n+1) /(4n+2))dx  =−(1/((4n+2)^2 )) ⇒∫_0 ^1  ((xlnx)/(1+x^4 ))dx =−Σ_(n=0) ^∞  (((−1)^n )/(4(2n+1)^2 ))  =−(k/4)( katalan constant)  ∫_0 ^1  ((xln(x+1))/(1+x^4 ))dx =∫_0 ^1 xln(x+1)Σ_(n=0) ^∞ (−1)^n  x^(4n)  dx  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^1  x^(4n+1)  ln(x+1)dx =Σ_(n=0) ^∞ (−1)^n  v_n   v_n =∫_0 ^1  x^(4n+1) ln(x+1)dx =_(x+1=t)   ∫_1 ^2  (t−1)^(4n+1) ln(t)dt  =∫_1 ^2  ln(t)Σ_(k=0) ^(4n+1)  C_(4n+1) ^k  t^k (−1)^(4n+1−k)  dt  =−Σ_(k=0) ^(4n+1)  (−1)^k  C_(4n+1) ^k  ∫_1 ^2  t^(k ) ln(t)dt   and by psrts  ∫_1 ^2  t^k  ln(t)dt =[(1/(k+1))t^(k+1) ln(t)]_1 ^2 −∫_1 ^2  (t^(k+1) /(k+1))(dt/t)  =((2^(k+1) ln(2))/(k+1))−(1/(k+1))∫_1 ^2   t^k  dt =((2^(k+1) ln(2))/(k+1))−(1/((k+1)^2 ))[t^(k+1) ]_1 ^2   =((2^(k+1) ln(2))/(k+1))−(1/((k+1)^2 ))(2^(k+1) −1) ⇒  v_n =−Σ_(k=0) ^(4n+1) (−1)^k  C_(4n+1) ^k {((2^(k+1) ln(2))/(k+1))−((2^(k+1) −1)/((k+1)^2 ))}  ....be continued...

A=0xln(1+x)1+x4dx=01xln(1+x)1+x4dx+1xln(x+1)1+x4dx1xln(x+1)1+x4dx=x=1t01ln(1+1t)t(1+1t4)(dtt2)=01ln(t+1)ln(tt3+1tdt=01tln(t+1)tlnt1+t4dtA=201xln(x+1)1+x4dx01xlnx1+x4dxwehave01xlnx1+x4dx=01xln(x)n=0(1)nx4ndx=n=0(1)n01x4n+1ln(x)dx=n=0(1)nunun=01x4n+1ln(x)dx=[14n+2x4n+2ln(x)]0101x4n+14n+2dx=1(4n+2)201xlnx1+x4dx=n=0(1)n4(2n+1)2=k4(katalanconstant)01xln(x+1)1+x4dx=01xln(x+1)n=0(1)nx4ndx=n=0(1)n01x4n+1ln(x+1)dx=n=0(1)nvnvn=01x4n+1ln(x+1)dx=x+1=t12(t1)4n+1ln(t)dt=12ln(t)k=04n+1C4n+1ktk(1)4n+1kdt=k=04n+1(1)kC4n+1k12tkln(t)dtandbypsrts12tkln(t)dt=[1k+1tk+1ln(t)]1212tk+1k+1dtt=2k+1ln(2)k+11k+112tkdt=2k+1ln(2)k+11(k+1)2[tk+1]12=2k+1ln(2)k+11(k+1)2(2k+11)vn=k=04n+1(1)kC4n+1k{2k+1ln(2)k+12k+11(k+1)2}....becontinued...

Terms of Service

Privacy Policy

Contact: info@tinkutara.com