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

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

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integral of 2+1/(1-x)

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

2 x - ln ∣ 1 - x ∣ + C

Solution steps

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

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

\int 2 d x = 2 x

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

= 2 x - ln ∣ 1 - x ∣

Add a constant to the solution

= 2 x - ln ∣ 1 - x ∣ + C

\frac{\partial}{\partial z} (z)

\frac{3}{( x - 1 )}

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

y = - 2 x - 1

n = 1 \sum \infty \frac{1}{n + 3 ^{n}}