What is the integral from 1 to 7 of 2/(4x-1) ?

The integral from 1 to 7 of 2/(4x-1) is ln(3)

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integral from 1 to 7 of 2/(4x-1)

\int_{1}^{7} \frac{2}{4 x - 1} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{1}^{7} \frac{2}{4 x - 1} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{7} \frac{1}{4 x - 1} d x

Apply u-substitution

: \int_{3}^{27} \frac{1}{4 u} d u

= 2 \cdot \int_{3}^{27} \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_{3}^{27} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{4} [ln ∣ u ∣]_{3}^{27}

Simplify

2 \cdot \frac{1}{4} [ln ∣ u ∣]_{3}^{27} : \frac{1}{2} [ln ∣ u ∣]_{3}^{27}

= \frac{1}{2} [ln ∣ u ∣]_{3}^{27}

Compute the boundaries:

2 ln (3)

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

Simplify

= ln (3)

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What is the integral from 1 to 7 of 2/(4x-1) ?
The integral from 1 to 7 of 2/(4x-1) is ln(3)