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)

\int_{3}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{\partial}{\partial y} ((x + 2 y + 3 z)^{\frac{3}{2}})

(1 + x^{2}) \frac{d y}{d x} - 2 x y = x^{2} + x^{4}

\frac{\partial}{\partial x} (20 e^{5 x} y)

f (x) = ln (x^{2} - 1)

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)