What is the integral of 1/(1+29x) ?

The integral of 1/(1+29x) is 1/29 ln|1+29x|+C

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integral of 1/(1+29x)

\int \frac{1}{1 + 29 x} d x

\frac{1}{29} ln ∣ 1 + 29 x ∣ + C

Solution steps

\int \frac{1}{1 + 29 x} d x

Apply u-substitution

: \int \frac{1}{29 u} d u

= \int \frac{1}{29 u} d u

Take the constant out:

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

= \frac{1}{29} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{29} ln ∣ u ∣

Substitute back

u = 1 + 29 x

= \frac{1}{29} ln ∣ 1 + 29 x ∣

Add a constant to the solution

= \frac{1}{29} ln ∣ 1 + 29 x ∣ + C

x \to 0 lim (- \frac{3}{x})

- \frac{20 , 0}{1 + 0.2 t} + 21, 0

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What is the integral of 1/(1+29x) ?
The integral of 1/(1+29x) is 1/29 ln|1+29x|+C