Question

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

Answer 1

`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)`

Answer 2

`(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,