What is the integral from 1 to 8 of 1/x-1/(x^2) ?

The integral from 1 to 8 of 1/x-1/(x^2) is 3ln(2)-7/8

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integral from 1 to 8 of 1/x-1/(x^2)

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

3 ln (2) - \frac{7}{8}

Decimal

1.20444 \dots

Solution steps

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

\int_{1}^{8} \frac{1}{x} d x = 3 ln (2)

\int_{1}^{8} \frac{1}{x ^{2}} d x = \frac{7}{8}

= 3 ln (2) - \frac{7}{8}

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What is the integral from 1 to 8 of 1/x-1/(x^2) ?
The integral from 1 to 8 of 1/x-1/(x^2) is 3ln(2)-7/8