What is the integral of 1/(0.15p+20) ?

The integral of 1/(0.15p+20) is 6.66666…ln|0.15p+20|+C

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integral of 1/(0.15p+20)

\int \frac{1}{0.15 p + 20} d p

6.66666 \dots ln ∣ 0.15 p + 20 ∣ + C

Solution steps

\int \frac{1}{0.15 p + 20} d p

Apply u-substitution

: \int \frac{1}{0.15 u} d u

= \int \frac{1}{0.15 u} d u

Take the constant out:

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

= \frac{1}{0.15} \cdot \int \frac{1}{u} d u

Simplify

= 6.66666 \dots \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6.66666 \dots ln ∣ u ∣

Substitute back

u = 0.15 p + 20

= 6.66666 \dots ln ∣ 0.15 p + 20 ∣

Add a constant to the solution

= 6.66666 \dots ln ∣ 0.15 p + 20 ∣ + C

\int tan^{2} (x) cos (x) d x

\frac{\partial}{\partial x} (b cos (w x) + cs in (w x))

n = 0 \sum \infty (\frac{- 6 + 7}{2})^{n}

g (x) = - x^{2} + 4 x^{2} - 2 x + 2

x^{2} - 4 x, 2 x + 7

What is the integral of 1/(0.15p+20) ?
The integral of 1/(0.15p+20) is 6.66666…ln|0.15p+20|+C