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

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

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

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

1 + ln (2)

Decimal

1.69314 \dots

Solution steps

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

Expand

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

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

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

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

= 1 + ln (2)

\int_{- \frac{π}{3}}^{0} sin (x) d x

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\int_{0}^{1} 1 + t^{2} d t

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\int_{- 2}^{4} 6 + 4 x - x^{2} d x

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