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

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

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integral of 1/(-x+1)

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

- ln ∣ - x + 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = - x + 1

= - ln ∣ - x + 1 ∣

Add a constant to the solution

= - ln ∣ - x + 1 ∣ + C

\frac{d}{d x} (\frac{x ^{2} + 4}{2 x})

y = 4 - 2 x^{2}

u + 1

\frac{\partial}{\partial x} (x^{3} + y^{5} + x^{7} - y^{2})

\frac{\partial}{\partial x} (A x)