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

Mar 8, 2018

$- 2 \ln | 1 - x | - x + C$

#### Explanation:

Given: $\int \frac{1 + x}{1 - x} \setminus \mathrm{dx}$

We know that $\frac{1 + x}{1 - x} = \frac{2 - \left(1 - x\right)}{1 - x}$

$= \frac{2}{1 - x} - 1$

Now, we got $\int \left(\frac{2}{1 - x} - 1\right) \setminus \mathrm{dx}$

Using the sum rule, $\int \left(f \left(a\right) + f \left(b\right)\right) \setminus \mathrm{dx} = \int f \left(a\right) + \int f \left(b\right)$

So, we have

$= \int \frac{2}{1 - x} \setminus \mathrm{dx} - \int 1 \setminus \mathrm{dx}$

Now, we use u-substitution to compute $\int \frac{2}{1 - x} \setminus \mathrm{dx}$.

Let $u = 1 - x$, then $\frac{\mathrm{du}}{\mathrm{dx}} = - 1 , \mathrm{dx} = \frac{\mathrm{du}}{-} 1 = - \mathrm{du}$.

$= \int \frac{2}{u} \cdot - \mathrm{du} - \int 1 \setminus \mathrm{dx}$

$= - \int \frac{2}{u} \setminus \mathrm{du} - x$

$= - 2 \int \frac{1}{u} \setminus \mathrm{du} - x$

$= - 2 \ln | u | - x$

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

$= - 2 \ln | 1 - x | - x$

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

$\textcolor{b l u e}{\overline{\underline{| - 2 \ln | 1 - x | - x + C |}}}$.