How do you evaluate the definite integral #intx/(x^2+1)# from #[0,sqrt(e-1)]#?

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Answer 1 · 2017-01-17T23:45:32

#int_0^sqrt(e-1)x/(x^2+1)dx=1/2#

#I=int_0^sqrt(e-1)x/(x^2+1)dx#

Let:

#u=x^2+1#
#du=2xcolor(white).dx#

When we perform this substitution, we will also change the current bounds:

#x=sqrt(e-1)" "=>" "u=x^2+1=(sqrt(e-1))^2+1=e#

#x=0" "=>" "u=x^2+1=0^2+1=1#

Then:

#I=1/2int_0^sqrt(e-1)(2x)/(x^2+1)dx=1/2int_1^e1/udu#

This is the integral of the natural logarithm:

#I=1/2(lnabsu)|_1^e=1/2(lne-ln1)=1/2(1-0)=1/2#