What is the integral from 0 to 1 of 2/(4x+5) ?

The integral from 0 to 1 of 2/(4x+5) is ln(3)-1/2 ln(5)

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integral from 0 to 1 of 2/(4x+5)

\int_{0}^{1} \frac{2}{4 x + 5} d x

ln (3) - \frac{1}{2} ln (5)

Decimal Notation

0.29389 \dots

Solution steps

\int_{0}^{1} \frac{2}{4 x + 5} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} \frac{1}{4 x + 5} d x

Apply u-substitution

: \int_{5}^{9} \frac{1}{4 u} d u

= 2 \cdot \int_{5}^{9} \frac{1}{4 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int_{5}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{4} [ln ∣ u ∣]_{5}^{9}

Simplify

2 \cdot \frac{1}{4} [ln ∣ u ∣]_{5}^{9} : \frac{1}{2} [ln ∣ u ∣]_{5}^{9}

= \frac{1}{2} [ln ∣ u ∣]_{5}^{9}

Compute the boundaries:

2 ln (3) - ln (5)

= \frac{1}{2} (2 ln (3) - ln (5))

Simplify

= ln (3) - \frac{1}{2} ln (5)

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What is the integral from 0 to 1 of 2/(4x+5) ?
The integral from 0 to 1 of 2/(4x+5) is ln(3)-1/2 ln(5)