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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{1} \frac{4}{2 x + 3} d x

2 (ln (5) - ln (3))

Decimal Notation

1.02165 \dots

Solution steps

\int_{0}^{1} \frac{4}{2 x + 3} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} \frac{1}{2 x + 3} d x

Apply u-substitution

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

= 4 \cdot \int_{3}^{5} \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{3}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{5}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{5} : 2 [ln ∣ u ∣]_{3}^{5}

= 2 [ln ∣ u ∣]_{3}^{5}

Compute the boundaries:

ln (5) - ln (3)

= 2 (ln (5) - ln (3))

\int_{0}^{a} y a^{2} - y^{2} d y

\int_{\frac{π}{6}}^{\frac{5 π}{6}} sin (x) d x

\int_{- 3}^{3} (3 x^{3} - 2 x^{2} + x + 1) d x

\int_{- \infty}^{4} 3 e^{- 3 x} d x

\int_{- 2}^{2} (2 - 4 - x^{2}) d x

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