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

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

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{9} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{2} [ln ∣ x ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

= \frac{1}{2} \cdot 2 ln (3)

Simplify

= ln (3)

\frac{d}{d x} (7 (2 x e^{x} + e^{x} x^{2}))

\int y + x e^{y} d y

x \to \infty lim (1 \cdot x)

\frac{d}{d x} (\frac{x ^{3} - 8}{x ^{3} + 8})

\frac{\partial}{\partial x} (x^{2} y - y^{2} - 7)

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