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

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

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

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

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

Decimal

0.80471 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{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_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

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

x \to 9 lim (\frac{x - 3}{x - 9})

25 x^{2}, 9

(6.2) (9.3)

n = 1 \sum \infty (cos (2.8))^{n}

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

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