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

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

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

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

y ln (2)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 y \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

= 2 y \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 y \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Simplify

2 y \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : y [ln ∣ u ∣]_{1}^{2}

= y [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= y ln (2)

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\int_{0}^{1} \frac{9}{1 + t ^{2}} d t

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