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

The integral from 2 to 3 of (1/(t-1)+t) is ln(2)+5/2

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

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

ln (2) + \frac{5}{2}

Decimal

3.19314 \dots

Solution steps

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

\int_{2}^{3} \frac{1}{t - 1} d t = ln (2)

\int_{2}^{3} t d t = \frac{5}{2}

= ln (2) + \frac{5}{2}

What is the integral from 2 to 3 of (1/(t-1)+t) ?
The integral from 2 to 3 of (1/(t-1)+t) is ln(2)+5/2