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

The integral from 0 to 3 of x/(2+x^2) is 1/2 (ln(11)-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 3 of x/(2+x^2)

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

\frac{1}{2} (ln (11) - ln (2))

Decimal

0.85237 \dots

Solution steps

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

Apply u-substitution

= \int_{2}^{11} \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_{2}^{11} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{11}

Compute the boundaries:

ln (11) - ln (2)

= \frac{1}{2} (ln (11) - ln (2))

What is the integral from 0 to 3 of x/(2+x^2) ?
The integral from 0 to 3 of x/(2+x^2) is 1/2 (ln(11)-ln(2))