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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Decimal

0.11157 \dots

Solution steps

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

Apply u-substitution

= \int_{4}^{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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{4}^{5}

Compute the boundaries:

ln (5) - 2 ln (2)

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

Simplify

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

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