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

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Answer 1 · 2016-08-24T13:10:44

#.191#

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

using u substitution let #u=ln(x^2+1)# then
#(du)/dx = 2x/(x^2+1)#
#1/2du= x/(x^2+1) dx #

#int 1/u 1/2du#

#1/2[lnu]#

now replace the u in terms of x

#1/2[ln(ln(x^2+1))]_1^2#

#1/2[ln(ln(2^2+1)) - ln(ln(1^2+1))] = .191#