What is the integral from 1 to 9 of (2x+1/x) ?

The integral from 1 to 9 of (2x+1/x) is 80+2ln(3)

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integral from 1 to 9 of (2x+1/x)

\int_{1}^{9} (2 x + \frac{1}{x}) d x

80 + 2 ln (3)

Decimal Notation

82.19722 \dots

Solution steps

\int_{1}^{9} 2 x + \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_{1}^{9} 2 x d x + \int_{1}^{9} \frac{1}{x} d x

\int_{1}^{9} 2 x d x = 80

\int_{1}^{9} \frac{1}{x} d x = 2 ln (3)

= 80 + 2 ln (3)

\int_{0}^{2} \frac{x ^{2}}{x ^{3} + 1} d x

\int_{0}^{4} \frac{1}{9 - 2 x} d x

\int_{0}^{x} \frac{1}{x ^{2} + y ^{2}} d y

\int_{- 2}^{2} 4 - y^{2} d y

\int_{0}^{1} - x^{2} + 1 d x

What is the integral from 1 to 9 of (2x+1/x) ?
The integral from 1 to 9 of (2x+1/x) is 80+2ln(3)