What is the integral of 1/(0.02y-6) ?

The integral of 1/(0.02y-6) is 50ln|0.02y-6|+C

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integral of 1/(0.02y-6)

\int \frac{1}{0.02 y - 6} d y

50 ln ∣ 0.02 y - 6 ∣ + C

Solution steps

\int \frac{1}{0.02 y - 6} d y

Apply u-substitution

: \int \frac{1}{0.02 u} d u

= \int \frac{1}{0.02 u} d u

Take the constant out:

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

= \frac{1}{0.02} \cdot \int \frac{1}{u} d u

Simplify

= 50 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 50 ln ∣ u ∣

Substitute back

u = 0.02 y - 6

= 50 ln ∣ 0.02 y - 6 ∣

Add a constant to the solution

= 50 ln ∣ 0.02 y - 6 ∣ + C

f (x) = x^{- 1}

\int_{0}^{2 π} \frac{1 - cos ( x )}{2} d x

\int \frac{x ^{2} - 16}{x ^{4}} d x

n = 1 \sum \infty \frac{7}{n ( n + 1 )}

x \to 1 + lim (\frac{1}{x ^{1 - x}})

What is the integral of 1/(0.02y-6) ?
The integral of 1/(0.02y-6) is 50ln|0.02y-6|+C