What is the integral of (2900)/(1+0.25t) ?

The integral of (2900)/(1+0.25t) is 11600ln|1+0.25t|+C

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integral of (2900)/(1+0.25t)

\int \frac{2900}{1 + 0.25 t} d t

11600 ln ∣ 1 + 0.25 t ∣ + C

Solution steps

\int \frac{2900}{1 + 0.25 t} d t

Take the constant out:

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

= 2900 \cdot \int \frac{1}{1 + 0.25 t} d t

Apply u-substitution

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

= 2900 \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

= 2900 \cdot \frac{1}{0.25} \cdot \int \frac{1}{u} d u

Simplify

= 2900 \cdot 4 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2900 \cdot 4 ln ∣ u ∣

Substitute back

u = 1 + 0.25 t

= 2900 \cdot 4 ln ∣ 1 + 0.25 t ∣

Simplify

= 11600 ln ∣ 1 + 0.25 t ∣

Add a constant to the solution

= 11600 ln ∣ 1 + 0.25 t ∣ + C

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What is the integral of (2900)/(1+0.25t) ?
The integral of (2900)/(1+0.25t) is 11600ln|1+0.25t|+C