What is the integral of (3100)/(1+0.25x) ?

The integral of (3100)/(1+0.25x) is 12400ln|1+0.25x|+C

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integral of (3100)/(1+0.25x)

\int \frac{3100}{1 + 0.25 x} d x

12400 ln ∣ 1 + 0.25 x ∣ + C

Solution steps

\int \frac{3100}{1 + 0.25 x} d x

Take the constant out:

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

= 3100 \cdot \int \frac{1}{1 + 0.25 x} d x

Apply u-substitution

: \int \frac{1}{0.25 u} d u

= 3100 \cdot \int \frac{1}{0.25 u} d u

Take the constant out:

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

= 3100 \cdot \frac{1}{0.25} \cdot \int \frac{1}{u} d u

Simplify

= 3100 \cdot 4 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3100 \cdot 4 ln ∣ u ∣

Substitute back

u = 1 + 0.25 x

= 3100 \cdot 4 ln ∣ 1 + 0.25 x ∣

Simplify

= 12400 ln ∣ 1 + 0.25 x ∣

Add a constant to the solution

= 12400 ln ∣ 1 + 0.25 x ∣ + C

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What is the integral of (3100)/(1+0.25x) ?
The integral of (3100)/(1+0.25x) is 12400ln|1+0.25x|+C