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

2 Answers
Ratnaker Mehta
Jun 3, 2018

# -ln(x+1)/x+ln|x/(x+1)|+C, or, #

#ln{|x*(x+1)^(-(1/x+1))|}+C#

Explanation:

The Rule of Integration by Parts :

#intuv'dx=uv-intu'vdx#.

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

We take, #u=ln(x+1) and v'=1/x^2#.

#:. u'=1/(x+1)and v=intv'dx=int1/x^2dx=-1/x#.

#:. I=-1/xln(x+1)-int{(1/(x+1))(-1/x)}dx#,

#=-ln(x+1)/x+int1/{x(x+1)}dx#,

#=-ln(x+1)/x+int{(x+1)-x}/{x(x+1)}dx#,

#=-ln(x+1)/x+int{(x+1)/{x(x+1)}-x/{x(x+1)}}dx#,

#=-ln(x+1)/x+int{1/x-1/(x+1)}dx#,

#=-ln(x+1)/x+{ln|x|-ln|(x+1)|}#,

#=-ln(x+1)/x+ln|x/(x+1)|#.

# rArr I=-ln(x+1)/x+ln|x/(x+1)|+C,#

#or, I=-1/xln|x+1|+ln|x/(x+1)|#,

#=ln(x+1)^(-1/x)+ln|x/(x+1)|#,

#=ln{|(x+1)^(-1/x)*x/(x+1)|}#.

#:. I=ln{|x*(x+1)^(-(1/x+1))|}+C#.

Enjoy Maths.!

Sonnhard
Jun 3, 2018

#-log(x+1)/x+log|x|-log|x+1|+C#

Explanation:

We use Integration by parts:
#int fdg=fg-intgdf#

with
#f=log(x+1),dg=1/x^2dx#
we get
#df=1/(x+1)dx,g=-1/x#
so we have
#-log(x+1)/x+int1/(x*(x+1))dx#
#=log(x+1)/x+int1/xdx-int1/(x+1)dx#
so we get
#-log(x+1)/x+log|x|-log|x+1|+C#

Related questions
See all questions in Integration by Parts
Impact of this question
1969 views around the world
You can reuse this answer
Creative Commons License