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

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

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

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

Decimal

0.34657 \dots

Solution steps

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

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

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