How do you integrate #(1+x)/(1-x)#?

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Answer 1 · 2018-03-08T12:36:28

#-2ln|1-x|-x+C#

Given: #int(1+x)/(1-x) \ dx#

We know that #(1+x)/(1-x)=(2-(1-x))/(1-x)#

#=2/(1-x)-1#

Now, we got #int(2/(1-x)-1) \ dx#

Using the sum rule, #int(f(a)+f(b)) \ dx=intf(a)+intf(b)#

So, we have

#=int2/(1-x) \ dx-int1 \ dx#

Now, we use u-substitution to compute #int2/(1-x) \ dx#.

Let #u=1-x#, then #(du)/(dx)=-1,dx=(du)/-1=-du#.

#=int2/u*-du-int1 \ dx#

#=-int2/u \ du-x#

#=-2int1/u \ du-x#

#=-2ln|u|-x#

Reversing the substitution that #u=1-x#, we get

#=-2ln|1-x|-x#

Now, we add a constant, and our final answer is

#color(blue)(barul(|-2ln|1-x|-x+C|))#.