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Question Number 67246 by Learner-123 last updated on 24/Aug/19

Integrate:  1) ∫_3 ^( ∞)  ((1/x dx)/(ln(x)(√(ln^2 x−1))))  2) ∫_1 ^∞ ((e^x dx)/(1+e^(2x) ))  3) ∫_1 ^∞  ((2^x dx)/(x+1))  4) ∫_2 ^∞  ((√x)/(ln(x)))dx

Integrate:1)31/xdxln(x)ln2x12)1exdx1+e2x3)12xdxx+14)2xln(x)dx

Commented by MJS last updated on 24/Aug/19

3) and 4) are not possible to solve with  elementary functions...

3)and4)arenotpossibletosolvewithelementaryfunctions...

Commented by mathmax by abdo last updated on 24/Aug/19

2) let I =∫_1 ^∞  (e^x /(1+e^(2x) ))dx   changement e^x  =t give  I =∫_e ^(+∞)  (t/(1+t^2 ))(dt/t) =∫_e ^(+∞)  (dt/(1+t^2 )) =[arctan(t)]_e ^(+∞)  =(π/2) −arctan(e).

2)letI=1ex1+e2xdxchangementex=tgiveI=e+t1+t2dtt=e+dt1+t2=[arctan(t)]e+=π2arctan(e).

Commented by mathmax by abdo last updated on 24/Aug/19

3) let I =∫_1 ^(+∞)  (2^x /(x+1))dx  changement  x=(1/t) give  I =−∫_0 ^1   (2^(1/t) /((1/t)+1)) (−(dt/t^2 )) =∫_0 ^1  (2^(1/t) /(t+t^2 )) dt  =∫_0 ^1  (2^(1/t) /(t(1+t)))dt  but we see that lim_(t→+∞)    (2^x /(x+1)) =+∞  so this integral diverges.

3)letI=1+2xx+1dxchangementx=1tgiveI=0121t1t+1(dtt2)=0121tt+t2dt=0121tt(1+t)dtbutweseethatlimt+2xx+1=+sothisintegraldiverges.

Commented by mathmax by abdo last updated on 24/Aug/19

4) changement (√x)=t give ∫_2 ^(+∞)  ((√x)/(lnx))dx =∫_(√2) ^(+∞)  (t/(ln(t^2 )))(2t)dt  = ∫_(√2) ^(+∞)   (t^2 /(ln(t)))dt     we have lim_(t→+∞)    (t^2 /(ln(t))) =+∞ so this integral  diverges .

4)changementx=tgive2+xlnxdx=2+tln(t2)(2t)dt=2+t2ln(t)dtwehavelimt+t2ln(t)=+sothisintegraldiverges.

Answered by MJS last updated on 24/Aug/19

1)  ∫(((1/x)dx)/(ln x (√(ln^2  x −1))))=       [t=ln x → dx=xdt]  =∫(dt/(t(√(t^2 −1))))=       [u=(√(t^2 −1)) → dt=((√(t^2 −1))/t)du]  =∫(du/(u^2 +1))=arctan u =arctan (√(t^2 −1)) =arctan (√(ln^2  x −1)) +C

1)1xdxlnxln2x1=[t=lnxdx=xdt]=dttt21=[u=t21dt=t21tdu]=duu2+1=arctanu=arctant21=arctanln2x1+C

Commented by MJS last updated on 24/Aug/19

in one step:  ∫(((1/x)dx)/(ln x (√(ln^2  x −1))))=       [t=(√(ln^2  x −1)) → dx=((x(√(ln^2  x −1)))/(ln x))]  =∫(dt/(t^2 +1))=...

inonestep:1xdxlnxln2x1=[t=ln2x1dx=xln2x1lnx]=dtt2+1=...

Commented by Learner-123 last updated on 25/Aug/19

thank you sir.

thankyousir.

Answered by MJS last updated on 24/Aug/19

2)  ∫(e^x /(1+e^(2x) ))dx=       [t=e^x  → dx=(dt/e^x )]  =∫(dt/(t^2 +1))=arctan t =arctan e^x  +C

2)ex1+e2xdx=[t=exdx=dtex]=dtt2+1=arctant=arctanex+C

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