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

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

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

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

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

Decimal Notation

0.34657 \dots

Solution steps

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

Apply u-substitution

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

= \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)

4 cos (x)

n = 1 \sum \infty (- 1)^{n} n^{- 2}

\frac{1}{( s + 1 ) ^{2} ( s ^{2} + 1 )}

(x 4 - x^{2})^{^{'}}

n = 1 \sum \infty \frac{1 ^{n}}{n ^{7}}

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