How do you evaluate the integral of int (2 -1/x)dx∫(2−1x)dx from 1/2 to 3?

1 Answer
Andrea S.
May 11, 2018

int_(1/2)^3 (2-1/x)dx =5 - ln6∫312(2−1x)dx=5−ln6

Explanation:

Using the linearity of the integral:

int_(1/2)^3 (2-1/x)dx = 2int_(1/2)^3dx - int_(1/2)^3 dx/x∫312(2−1x)dx=2∫312dx−∫312dxx

int_(1/2)^3 (2-1/x)dx = 2(3-1/2) - (ln3-ln(1/2))∫312(2−1x)dx=2(3−12)−(ln3−ln(12))

int_(1/2)^3 (2-1/x)dx =5 - ln6∫312(2−1x)dx=5−ln6

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