Understanding : $\int \frac{1}{x^{-1}+1}dx$

Question

I am trying to practice integrating the following: $$\int \frac{1}{x^{-1}+1}dx$$

The first step of simplifying the fraction does not make sense to me in how it becomes: $$-\frac{1}{1+x}+1 $$

Could someone please explain how this step is done?

Answer 1

Note:

$$\frac{1}{x^{-1} + 1}$$ $$ = \frac{1}{\frac{1}{x} + 1}$$ $$ = \frac{1}{\frac{1 + x}{x}}$$ $$ = \frac{x}{1 + x}$$ $$ = 1 - \frac{1}{1 + x}$$

Answer 2

$$ \frac{1}{x^{-1}+1} = \frac{x}{1+x} \\ = \frac{1+x-1}{1+x}\\ = 1-\frac{1}{1+x}\\ = -\frac{1}{1+x}+1\\ $$

Answer 3

Combining the other two answers, we can solve the integral:

\begin{align} &\quad\int\frac{1}{x^{-1}+1}\ dx\\ &=\int\frac{1}{\frac{1}{x}+1}\ dx\\ &=\int\frac{1}{\frac{x+1}{x}}\ dx\\ &=\int\frac{x}{x+1}\ dx \end{align}

Now, by $u$-sub, we can apply the substitution $u=x+1$.

\begin{align} \frac{du}{dx}&=1\\ du&=dx \end{align}

We now get

\begin{align} &\quad\int\frac{x}{x+1}\ dx\\ &=\int\frac{u-1}{u}\ du\\ &=\int1-u^{-1}\ du\\ &=u-\ln|u|+c\\ &=(x+1)-\ln|x+1|+c \end{align}