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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln (3)

Decimal

1.09861 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int_{2}^{6} \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{2}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{6}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{6} : [ln ∣ u ∣]_{2}^{6}

= [ln ∣ u ∣]_{2}^{6}

Compute the boundaries:

ln (6) - ln (2)

= ln (6) - ln (2)

Simplify

= ln (3)

\int_{0}^{2} x^{3} (2 - x) d x

\int_{e}^{\infty} \frac{1}{x} d x

\int_{0}^{3 t} \frac{sin ( u )}{u} d u

\int_{0}^{\infty} x \cdot e^{x} d x

\int_{0}^{π} x sin (k x) d x

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