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

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

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integral from 1 to 2 of (3-x)-(2/x)

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

- 2 ln (2) + \frac{3}{2}

Decimal

0.11370 \dots

Solution steps

\int_{1}^{2} 3 - 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} 3 d x - \int_{1}^{2} x d x - \int_{1}^{2} \frac{2}{x} d x

\int_{1}^{2} 3 d x = 3

\int_{1}^{2} x d x = \frac{3}{2}

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

= 3 - \frac{3}{2} - 2 ln (2)

Combine the fractions

3 - \frac{3}{2} : \frac{3}{2}

= - 2 ln (2) + \frac{3}{2}

\int_{- 1}^{1} x^{\frac{3}{2}} d x

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

\int_{1}^{4} \frac{8}{x ^{2}} d x

\int_{0}^{1} x^{\frac{1}{7}} d x

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

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