What is the integral of $\int (2 - \frac{1}{x}) d x$ $int (2 -1/x)dx$ from 1/2 to 3?

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What is the integral of $\int (2 - \frac{1}{x}) d x$ $int (2 -1/x)dx$ from 1/2 to 3?

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Answer 1 · 2016-09-23T03:19:15

$5 - ln (6)$ $5-ln(6)$

Explanation:

Integrate each of these by part. Note that $\frac{d}{d x} (2 x) = 2$ $d/dx(2x)=2$ , so $\int 2 d x = 2 x$ $int2dx=2x$ , and $\frac{d}{d x} (ln (x)) = \frac{1}{x}$ $d/dx(ln(x))=1/x$ , so $\int \frac{1}{x} d x = ln (x)$ $int1/xdx=ln(x)$ :

$\int_{\frac{1}{2}}^{3} (2 - \frac{1}{x}) d x = {[2 x - ln (x)]}_{\frac{1}{2}}^{3}$ $int_(1/2)^3(2-1/x)dx=[2x-ln(x)]_(1/2)^3$

$= [2 (3) - ln (3)] - [2 (\frac{1}{2}) - ln (\frac{1}{2})]$ $=[2(3)-ln(3)]-[2(1/2)-ln(1/2)]$

$= 6 - ln (3) - 1 + ln (\frac{1}{2})$ $=6-ln(3)-1+ln(1/2)$

Note that $- ln (3) + ln (\frac{1}{2}) = - ln (3) - ln (2) = - (ln (3) + ln (2)) = - ln (6)$ $-ln(3)+ln(1/2)=-ln(3)-ln(2)=-(ln(3)+ln(2))=-ln(6)$ :

$= 5 - ln (6)$ $=5-ln(6)$

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What is the integral of $\int (2 - \frac{1}{x}) d x$ $int (2 -1/x)dx$ from 1/2 to 3?

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