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

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

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

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

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

Decimal Notation

0.80685 \dots

Solution steps

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

\int_{1}^{2} x d x = \frac{3}{2}

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

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

\int_{2}^{5} \frac{ln ( x ^{5} )}{x} d x

(1 + x)^{\frac{1}{3}}

\int 5 e^{2 x} d x

y^{^{''}} + 9 y = t e^{3 t}

y^{^{'}} - y = 8 e^{x} + 18 e^{4 x}

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