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

The integral from 2 to 8 of (x^2+x+1+1/x) is 2ln(2)+204

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

\int_{2}^{8} (x^{2} + x + 1 + \frac{1}{x}) d x

2 ln (2) + 204

Decimal Notation

205.38629 \dots

Solution steps

\int_{2}^{8} x^{2} + x + 1 + \frac{1}{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_{2}^{8} x^{2} d x + \int_{2}^{8} x d x + \int_{2}^{8} 1 d x + \int_{2}^{8} \frac{1}{x} d x

\int_{2}^{8} x^{2} d x = 168

\int_{2}^{8} x d x = 30

\int_{2}^{8} 1 d x = 6

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

= 168 + 30 + 6 + 2 ln (2)

Simplify

= 2 ln (2) + 204

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What is the integral from 2 to 8 of (x^2+x+1+1/x) ?
The integral from 2 to 8 of (x^2+x+1+1/x) is 2ln(2)+204