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

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

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

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

2 ln (2) + \frac{3}{2}

Decimal

2.88629 \dots

Solution steps

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

Expand

(\frac{x + 1}{x})^{2} : 1 + \frac{2}{x} + \frac{1}{x ^{2}}

= \int_{1}^{2} 1 + \frac{2}{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}^{2} 1 d x + \int_{1}^{2} \frac{2}{x} d x + \int_{1}^{2} \frac{1}{x ^{2}} d x

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

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

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

= 1 + 2 ln (2) + \frac{1}{2}

Combine the fractions

1 + \frac{1}{2} : \frac{3}{2}

= 2 ln (2) + \frac{3}{2}

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