What is the integral from 1 to 2 of (cos(2x)-2/x) ?

The integral from 1 to 2 of (cos(2x)-2/x) is 1/2 (sin(4)-sin(2))-2ln(2)

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integral from 1 to 2 of (cos(2x)-2/x)

\int_{1}^{2} (cos (2 x) - \frac{2}{x}) d x

\frac{1}{2} (sin (4) - sin (2)) - 2 ln (2)

Decimal Notation

- 2.21934 \dots

Solution steps

\int_{1}^{2} cos (2 x) - \frac{2}{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

= \int_{1}^{2} cos (2 x) d x - \int_{1}^{2} \frac{2}{x} d x

\int_{1}^{2} cos (2 x) d x = \frac{1}{2} (sin (4) - sin (2))

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

= \frac{1}{2} (sin (4) - sin (2)) - 2 ln (2)

f (x) = (- 4 x^{2} + 9)^{2}

\frac{d}{d x} (\frac{12}{x + 1})

x \to - 3 + lim (x)

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

f (x) = \frac{2 x}{3}

What is the integral from 1 to 2 of (cos(2x)-2/x) ?
The integral from 1 to 2 of (cos(2x)-2/x) is 1/2 (sin(4)-sin(2))-2ln(2)