What is the integral from 1 to 4 of (sqrt(x)+1)/x ?

The integral from 1 to 4 of (sqrt(x)+1)/x is 2+2ln(2)

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integral from 1 to 4 of (sqrt(x)+1)/x

\int_{1}^{4} \frac{x + 1}{x} d x

2 + 2 ln (2)

Solution steps

\int_{1}^{4} \frac{x + 1}{x} d x

Expand

\frac{x + 1}{x} : \frac{1}{x} + \frac{1}{x}

= \int_{1}^{4} \frac{1}{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}^{4} \frac{1}{x} d x + \int_{1}^{4} \frac{1}{x} d x

\int_{1}^{4} \frac{1}{x} d x = 2

\int_{1}^{4} \frac{1}{x} d x = 2 ln (2)

= 2 + 2 ln (2)

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What is the integral from 1 to 4 of (sqrt(x)+1)/x ?
The integral from 1 to 4 of (sqrt(x)+1)/x is 2+2ln(2)