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

2 Answers
May 10, 2018

I=1/2ln|ln5/ln2|

or I~~0.42

Explanation:

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]}

May 10, 2018

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

Explanation:

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