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

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

Popular >

integral from 0 to 1 of 1/(3x+2)

\int_{0}^{1} \frac{1}{3 x + 2} d x

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

Decimal Notation

0.30543 \dots

Solution steps

\int_{0}^{1} \frac{1}{3 x + 2} d x

Apply u-substitution

: \int_{2}^{5} \frac{1}{3 u} d u

= \int_{2}^{5} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{2}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

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

\int_{0}^{16} \frac{x}{4 + 2 x} d x

\int_{7}^{14} 0.05 e^{- 0.05 x} d x

\int_{- \infty}^{0} e^{\frac{x}{4}} d x

\int_{1}^{5} (6 n^{- 2} - n^{- 3}) d n

\int_{0}^{π} \frac{1}{1 + sin ^{2} ( x )} d x

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