What is the integral of 1/((35-x)) ?

The integral of 1/((35-x)) is -ln|35-x|+C

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integral of 1/((35-x))

integral

\frac{1}{( 35 - x )}

- ln ∣ 35 - x ∣ + C

Solution steps

\int \frac{1}{35 - 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 = 35 - x

= - ln ∣ 35 - x ∣

Add a constant to the solution

= - ln ∣ 35 - x ∣ + C

\frac{1}{( x ^{2} - 4 x )}

(2 x + 3)^{4}

e \cdot x^{2}

e^{2 x^{2}}

x^{2} + 6 x