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

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

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integral of 1/(a-x)

integral

\frac{1}{a - x}

- ln ∣ a - x ∣ + C

Solution steps

\int \frac{1}{a - x} 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 = a - x

= - ln ∣ a - x ∣

Add a constant to the solution

= - ln ∣ a - x ∣ + C

(e^{x})^{- 1}

\frac{1}{x ( x + 1 )}

\frac{( x ^{2} + 2 x - 1 )}{( x ^{3} - x )}

ln (\frac{1}{x})

\frac{1}{( x ^{3} - 1 ) ^{2}}