What is the integral from 4 to 9 of (2+sqrt(t))/t ?

The integral from 4 to 9 of (2+sqrt(t))/t is 2(2ln(3)-2ln(2))+2

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integral from 4 to 9 of (2+sqrt(t))/t

\int_{4}^{9} \frac{2 + t}{t} d t

2 (2 ln (3) - 2 ln (2)) + 2

Decimal Notation

3.62186 \dots

Solution steps

\int_{4}^{9} \frac{2 + t}{t} d t

Expand

\frac{2 + t}{t} : \frac{2}{t} + \frac{1}{t}

= \int_{4}^{9} \frac{2}{t} + \frac{1}{t} 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_{4}^{9} \frac{2}{t} d t + \int_{4}^{9} \frac{1}{t} d t

\int_{4}^{9} \frac{2}{t} d t = 2 (2 ln (3) - 2 ln (2))

\int_{4}^{9} \frac{1}{t} d t = 2

= 2 (2 ln (3) - 2 ln (2)) + 2

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What is the integral from 4 to 9 of (2+sqrt(t))/t ?
The integral from 4 to 9 of (2+sqrt(t))/t is 2(2ln(3)-2ln(2))+2