What is the integral of 1/(100-x) ?

The integral of 1/(100-x) is -ln|100-x|+C

Popular >

integral of 1/(100-x)

\int \frac{1}{100 - x} d x

- ln ∣ 100 - x ∣ + C

Solution steps

\int \frac{1}{100 - 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 = 100 - x

= - ln ∣ 100 - x ∣

Add a constant to the solution

= - ln ∣ 100 - x ∣ + C

f (x) = \frac{6 x ^{4} - 9 x ^{2}}{3 x ^{3}}

\frac{d}{d x} (2 e^{3 x} - 2 x - 2)

\int_{1}^{2} (x^{2} - 5 x + 3) d x

\int \frac{5}{x 4 - ln ^{2} ( x )} d x

\frac{d}{d x} (- 12 x cos (x^{2}) + 8 x^{3} sin (x^{2}))