Integrate the following? lnx/x^2 dx

2 Answers
Mar 1, 2018

#-(lnx-1)/x+C#

Explanation:

We have:

#intlnx/x^2dx#

Which we can write as:

#intlnx*1/x^2dx#

Integration by parts states that:

#intuvdx#, where #u# and #v# are functions, is equal to:

#uintvdx-intu'(intvdx)dx#

Here, #u=lnx# and #v=1/x^2#

Inputting:

#lnx int1/x^2dx-int(d/dxlnx)(int1/x^2dx)dx#

#lnx(-1/x)-int(1/x)(-1/x)dx#

#-lnx/x-int-1/x^2dx#

#-lnx/x+int1/x^2dx#

#-lnx/x-1/x#

#-(lnx-1)/x#

Add the constant of integration:

#-(lnx-1)/x+C#

Mar 1, 2018

#-(lnx+1)/x+C#

Explanation:

Use integration by parts and differentiate # lnx# and integrate #1/x^2 #
as following:
the forumla for I.B.P is
#int f(x)g'(x)dx=[f(x)g(x)]-int f'(x)g(x)dx#
and let
#lnx=f(x) and 1/x^2 = g'(x )#
which means
#f'(x)=1/x and g(x)=-1/x #
all that is left now is using the I.B.P formula and solve a trivial integral, and simplify the answer. Hence,
#int lnx/x^2dx=[lnx * (-1/x) ]-int 1/x * (-1/x)dx #
simplifying gives
#=- lnx/x +int 1/x^2dx #
which yields
#=- lnx/x +(-1/x) #
which is the same as
#=- (lnx+1)/x #
dont forget to add the +C
#=- (lnx+1)/x+C #