What is the integral from 0 to 1 of (1+20t)/(1+4t) ?

The integral from 0 to 1 of (1+20t)/(1+4t) is 5-ln(5)

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integral from 0 to 1 of (1+20t)/(1+4t)

\int_{0}^{1} \frac{1 + 20 t}{1 + 4 t} d t

5 - ln (5)

Decimal

3.39056 \dots

Solution steps

\int_{0}^{1} \frac{1 + 20 t}{1 + 4 t} d t

Long division

\frac{1 + 20 t}{1 + 4 t} : 5 - \frac{4}{4 t + 1}

= \int_{0}^{1} 5 - \frac{4}{4 t + 1} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 5 d t - \int_{0}^{1} \frac{4}{4 t + 1} d t

\int_{0}^{1} 5 d t = 5

\int_{0}^{1} \frac{4}{4 t + 1} d t = ln (5)

= 5 - ln (5)

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What is the integral from 0 to 1 of (1+20t)/(1+4t) ?
The integral from 0 to 1 of (1+20t)/(1+4t) is 5-ln(5)