What is the integral from 1 to 3 of (2t+3)/(t+1) ?

The integral from 1 to 3 of (2t+3)/(t+1) is 4+ln(2)

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integral from 1 to 3 of (2t+3)/(t+1)

\int_{1}^{3} \frac{2 t + 3}{t + 1} d t

4 + ln (2)

Decimal Notation

4.69314 \dots

Solution steps

\int_{1}^{3} \frac{2 t + 3}{t + 1} d t

Apply u-substitution

: \int_{2}^{4} \frac{2 u + 1}{u} d u

= \int_{2}^{4} \frac{2 u + 1}{u} d u

Expand

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

= \int_{2}^{4} 2 + \frac{1}{u} d u

Apply the Sum Rule:

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

= \int_{2}^{4} 2 d u + \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} 2 d u = 4

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= 4 + ln (2)

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What is the integral from 1 to 3 of (2t+3)/(t+1) ?
The integral from 1 to 3 of (2t+3)/(t+1) is 4+ln(2)