Integrate #(1-x)/(1+x)^2#?

2 Answers
Sep 11, 2017

#int(1-x)/(1+x)^2dx=-ln|1+x|-2/(1+x)#

Explanation:

This is done by using partial fractions.

Let #(1-x)/(1+x)^2=A/(1+x)+B/(1+x)^2# and then we have

#1-x=A(1+x)+B=Ax+A+B#

and comparing coefficients on both sides, we get

#A=-1# and #B=2# and #(1-x)/(1+x)^2=-1/(1+x)+2/(1+x)^2#

Hence, #int(1-x)/(1+x)^2dx=-intdx/(1+x)+2intdx/(1+x)^2#

= #-ln|1+x|+(2(1+x)^(-1))/(-1)#

= #-ln|1+x|-2/(1+x)#

#(1-x)/(1+x)^2#
Seperate numerators
#1/((1+x)^2)-x/((x+1)^2)#
lets call these terms as,
#1/((1+x)^2)#=L
#x/((x+1)^2)#=M
,
Now we will evaluate #L#
this is a simple integration,
Integration of # x^n =x^(n+1)/(n+1)#
now,
#L=1/(1+x)^2#
#L=(1+x)^-2#
as #{(1/x^m)=x^(-m)}#

NOW as per above,

L=#(1+x)^(-2+1)/(-2+1)#
L=#(1+x)^-1/(-1)#
#L=-1/(1+x)#

Ahh,L is solved now :)!

Now lets deal with M,
we will solve M with Substitution,
#M=x/(x+1)^2#
now,lets take #(x+1)=t#

{Why not x=t???? Because taking x=t will not do any thing good to
M as M will show as #t/(1+t)^2# see nothing happened , so (1+x)=t}

Now as
(1+x)=t
diffrentiating,
#d(1+x)/dx=d(t)/(dx)#
#d(x)=d(t)#
Now,
M=#(t-1)/(t^2)*dt#
Seperating denominator as we did initially,
M=#(t/(t^2)-1/t^2)dt#

M=(#1/t-1/t^2#)dt
Integration of #1/x#=ln(x)
To integrate M,

M=#(ln(t)-(t^-(2+1))/(-2+1))#+C

M=#ln(t)-t^-1/(-1)#+C

M=#ln(t)+1/(t)#+C

Now to write the answer in tems of x
so,1+x=t
M=#ln(1+x)+1/(1+x)# +C

Add
L-M
YOu got the answer,