What is the integral from 1 to 4 of ((2x+1))/(2x) ?

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

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integral from 1 to 4 of ((2x+1))/(2x)

\int_{1}^{4} \frac{( 2 x + 1 )}{2 x} d x

3 + ln (2)

Decimal Notation

3.69314 \dots

Solution steps

\int_{1}^{4} \frac{2 x + 1}{2 x} d x

Take the constant out:

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

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

Expand

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

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

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_{1}^{4} 2 d x + \int_{1}^{4} \frac{1}{x} d x)

\int_{1}^{4} 2 d x = 6

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

= \frac{1}{2} (6 + 2 ln (2))

Simplify

\frac{1}{2} (6 + 2 ln (2)) : 3 + ln (2)

= 3 + ln (2)

\int sin (2 x) (3 - cos (2 x))^{2} d x

x \to 2 lim (\frac{1}{x ^{2} - 4 x + 4})

\frac{5}{s} - \frac{5}{s} e^{- 5 s}

n = 0 \sum \infty - (- 1)^{n + 1}

\frac{d}{d x} (\frac{3}{4 x ^{2}})

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