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

Question

Evelyn Agard · Accepted Answer

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

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

Christopher Agresta · Answer

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

Christopher Agresta · Answer

undefined To integrate $ \frac{\ln(x+1)}{x^2} $, you can use integration by parts. Let $ u = \ln(x+1) $ and $ dv = \frac{1}{x^2} \, dx $. Then, find $ du $ and $ v $, and apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

After integrating by parts, you'll get an expression in terms of $ x $ and $ \ln(x+1) $.