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

The integral from 0 to 1 of 1/(4x+3) is 1/4 (ln(7)-ln(3))

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integral from 0 to 1 of 1/(4x+3)

\int_{0}^{1} \frac{1}{4 x + 3} d x

\frac{1}{4} (ln (7) - ln (3))

Decimal

0.21182 \dots

Solution steps

\int_{0}^{1} \frac{1}{4 x + 3} d x

Apply u-substitution

: \int_{3}^{7} \frac{1}{4 u} d u

= \int_{3}^{7} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{3}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{3}^{7}

Compute the boundaries:

ln (7) - ln (3)

= \frac{1}{4} (ln (7) - ln (3))

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What is the integral from 0 to 1 of 1/(4x+3) ?
The integral from 0 to 1 of 1/(4x+3) is 1/4 (ln(7)-ln(3))