What is the integral from 0 to 1 of x 4/(pi(1+x^2)) ?

The integral from 0 to 1 of x 4/(pi(1+x^2)) is 2/pi ln(2)

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integral from 0 to 1 of x 4/(pi(1+x^2))

\int_{0}^{1} x \frac{4}{π ( 1 + x ^{2} )} d x

\frac{2}{π} ln (2)

Decimal

0.44127 \dots

Solution steps

\int_{0}^{1} x \frac{4}{π ( 1 + x ^{2} )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{4}{π} \cdot \int_{0}^{1} x \frac{1}{1 + x ^{2}} d x

Simplify

x \frac{1}{1 + x ^{2}} : \frac{x}{1 + x ^{2}}

= \frac{4}{π} \cdot \int_{0}^{1} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

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

= \frac{4}{π} \cdot \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{4}{π} \cdot \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{4}{π} \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Simplify

\frac{4}{π} \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : \frac{2}{π} [ln ∣ u ∣]_{1}^{2}

= \frac{2}{π} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= \frac{2}{π} ln (2)

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What is the integral from 0 to 1 of x 4/(pi(1+x^2)) ?
The integral from 0 to 1 of x 4/(pi(1+x^2)) is 2/pi ln(2)