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)

x y^{^{'}} + y = 55 x

\int \frac{5 sin ( 2 x )}{sin ( x )} d x

y = 1 0^{n x}

\int \frac{e ^{- 3 x}}{x} d x

\frac{d}{d x} (x^{2} \frac{1}{2})

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)