What is the integral from 1 to 2 of (e^{2t}+1/(2t)) ?

The integral from 1 to 2 of (e^{2t}+1/(2t)) is 1/2 (e^4-e^2+ln(2))

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integral from 1 to 2 of (e^{2t}+1/(2t))

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

\frac{1}{2} (e^{4} - e^{2} + ln (2))

Decimal

23.95112 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

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

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

= \frac{1}{2} (e^{4} - e^{2} + ln (2))

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What is the integral from 1 to 2 of (e^{2t}+1/(2t)) ?
The integral from 1 to 2 of (e^{2t}+1/(2t)) is 1/2 (e^4-e^2+ln(2))