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)

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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 Notation

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

= \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)

y^{4} x^{2} y^{^{'}} = 1

\frac{1}{( x + 4 ) ^{2}}

4 e^{x} cos (x)

ln (2 x) \cdot 4

\frac{d}{d x} (5^{- 4 x})

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)