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

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

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