How do you integrate #(ln(x+1)/(x^2)) dx#?

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Answer 1 · 2016-07-23T18:57:19

#= - (1/x + 1) ln(x+1) + ln x +C#

#int \ (ln(x+1)/(x^2)) \ dx#

#= int \ ln(x+1) d/dx( -1/x) \ dx#

by IBP this becomes:

#=- 1/x ln(x+1) + int \ d/dx (ln(x+1)) * 1/x \ dx#

#=- 1/x ln(x+1) + int \ 1/(x(x+1)) \ dx#

so some partial fractions on this integral

# 1/(x(x+1)) = A/ x + B/(x+1) = (A(x+1) + B x)/(x(x+1)) #

#x = 0, 1 = A#
#x = -1, 1 = -B#

#implies - 1/x ln(x+1) + int \ 1/ x - 1/(x+1)\ dx#

#= - 1/x ln(x+1) + ln x - ln (x+1) +C#

#= - (1/x + 1) ln(x+1) + ln x +C#