What is the integral of 2/(x^{-1)+1} ?

The integral of 2/(x^{-1)+1} is 2(-ln|1+x|+x)+C

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integral of 2/(x^{-1)+1}

\int \frac{2}{x ^{- 1} + 1} d x

2 (- ln ∣ 1 + x ∣ + x) + C

Solution steps

\int \frac{2}{x ^{- 1} + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{- 1} + 1} d x

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

= 2 \cdot \int - \frac{1}{1 + x} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int \frac{1}{1 + x} d x + \int 1 d x)

\int \frac{1}{1 + x} d x = ln ∣ 1 + x ∣

\int 1 d x = x

= 2 (- ln ∣ 1 + x ∣ + x)

Add a constant to the solution

= 2 (- ln ∣ 1 + x ∣ + x) + C

\int \frac{4}{x ^{2} - 4 x + 9} d x

(\frac{x ^{'}}{x})^{^{'}}

\int \frac{72}{w ^{2} 36 - w ^{2}} d w

y^{^{'}} + sin (t) y = sin (t), y (0) = 8

\frac{\partial}{\partial x} (3 x^{2} y - 4 x y^{2})

What is the integral of 2/(x^{-1)+1} ?
The integral of 2/(x^{-1)+1} is 2(-ln|1+x|+x)+C