What is the antiderivative of # (ln(x+1)/(x^2))#?

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Answer 1 · 2016-01-09T01:09:11

#I = -ln(x+1)/x + ln|x| - ln|x+1| + c#

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

Let's say #u = ln(x+1)# so #du = 1/(x+1)#, and #dv = 1/x^2# so #v = -1/x#

#I = -ln(x+1)/x + intdx/(x(x+1))#

The latter integral can only be solved with parcial fractions, so we assume there is a sum of fractions

#a/x + b/(x+1) = 1/(x(x+1))#

Where #a# and #b# are constants, that is also to say

#(a(x+1) + bx)/(x(x+1)) = 1/(x(x+1))#

Note that it means, that, for any value of #x#

#a(x+1) + bx = 1#

So if we assume #x = 0#,

#a(0+1) + b*0 = 1#
#a = 1#

And if we assume #x = -1#

#a(-1+1) -b = 1#
#-b = 1#
#b = -1#

So, back to the first integral we have

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

From there, it's an easy integral

#I = -ln(x+1)/x + ln|x| - ln|x+1| + c#