What is the integral of 1/(6-x) ?

The integral of 1/(6-x) is -ln|6-x|+C

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integral of 1/(6-x)

\int \frac{1}{6 - x} d x

- ln ∣ 6 - x ∣ + C

Solution steps

\int \frac{1}{6 - 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 = 6 - x

= - ln ∣ 6 - x ∣

Add a constant to the solution

= - ln ∣ 6 - x ∣ + C

6 π x^{2}

\int 14 x^{- 1} d x

\frac{d x}{d t} + 3 t x^{3} + \frac{x}{t} = 0

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

x y^{^{'}} + 2 y = 3 e^{x}