How do you find the definite integral of #(x dx) / [(x^2 +1)(ln (x^2 +1))]# from #[1, 2]#?

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Answer 1 · 2018-05-10T17:53:49

#I=1/2ln|ln5/ln2|#

# or I~~0.42#

We have,

#I=int_1^2x/((x^2+1)(ln(x^2+1)))dx#

Subst.

#ln(x^2+1)=u=>x^2+1=e^u=>2xdx=e^udu#

#=>xdx=1/2e^udu#

#x=1=>u=ln(1^2+1)=ln2=a, to (say)#

#x=2=>u=ln(2^2+1)=ln5=b, to (say)#

So,

#I=int_a^b (1/2e^u)/(e^u(u))du#

#I=1/2int_a^b 1/udu#

#=1/2[lnu]_a^b#

#=1/2[lnb-lna]#

#=1/2ln|b/a|,where,a=ln2 and b=ln5#

#I=1/2ln|ln5/ln2|#

#I~~0.42#

#f(x)=x/((x^2+1)(ln(x^2+1))# is continuous in #[1,2]#

graph{x/((x^2+1)(ln(x^2+1))) [-4, 6, -1.7, 3.3]}

Answer 2 · 2018-05-10T17:57:00

the answer
#int_1^2(x dx) / [(x^2 +1)(ln (x^2 +1))]=1/2[ln(ln(x^2+1))]_1^2=(ln(ln(5))-ln(ln(2)))/2=0.421198958#

show below

#int_1^2(x dx) / [(x^2 +1)(ln (x^2 +1))]#

suppose:

#u=x^2+1#

#du=2x*dx#

#dx=(du)/(2x)#

#x=sqrt(u-1)#

#int_1^2[sqrt(u-1)*(du)/(sqrt(u-1))]/[u*lnu]#

#1/2int_1^2[du]/[u*lnu]#

#1/2int_1^2[1/u*du]/[lnu]=1/2[ln(ln(u))]_1^2#

#1/2[ln(ln(x^2+1))]_1^2=(ln(ln(5))-ln(ln(2)))/2=0.421198958#