What is the integral from 0 to 1 of 1/(2x+4) ?

The integral from 0 to 1 of 1/(2x+4) is 1/2 ln(3/2)

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integral from 0 to 1 of 1/(2x+4)

\int_{0}^{1} \frac{1}{2 x + 4} d x

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

Decimal

0.20273 \dots

Solution steps

\int_{0}^{1} \frac{1}{2 x + 4} d x

Apply u-substitution

= \int_{4}^{6} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{4}^{6}

Compute the boundaries:

ln (6) - 2 ln (2)

= \frac{1}{2} (ln (6) - 2 ln (2))

Simplify

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

What is the integral from 0 to 1 of 1/(2x+4) ?
The integral from 0 to 1 of 1/(2x+4) is 1/2 ln(3/2)