What is the integral from 0 to 3 of (3000)/(1+0.25t) ?

The integral from 0 to 3 of (3000)/(1+0.25t) is 6715.38945…

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integral from 0 to 3 of (3000)/(1+0.25t)

\int_{0}^{3} \frac{3000}{1 + 0.25 t} d t

6715.38945 \dots

Solution steps

\int_{0}^{3} \frac{3000}{1 + 0.25 t} d t

Take the constant out:

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

= 3000 \cdot \int_{0}^{3} \frac{1}{1 + 0.25 t} d t

Apply u-substitution

= 3000 \cdot \int_{1}^{1.75} \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

= 3000 \cdot \frac{1}{0.25} \cdot \int_{1}^{1.75} \frac{1}{u} d u

Simplify

= 3000 \cdot 4 \cdot \int_{1}^{1.75} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3000 \cdot 4 [ln ∣ u ∣]_{1}^{1.75}

Simplify

= 12000 [ln ∣ u ∣]_{1}^{1.75}

Compute the boundaries:

0.55961 \dots

= 12000 \cdot 0.55961 \dots

Simplify

= 6715.38945 \dots

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What is the integral from 0 to 3 of (3000)/(1+0.25t) ?
The integral from 0 to 3 of (3000)/(1+0.25t) is 6715.38945…