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

The integral from 0 to 1 of x/(1+xy) is ln|x+1|

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

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

ln ∣ x + 1 ∣

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{1 + x} \frac{1}{xu} d u

= x \cdot \int_{1}^{1 + x} \frac{1}{xu} d u

Take the constant out:

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

= x \frac{1}{x} \cdot \int_{1}^{1 + x} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= x \frac{1}{x} [ln ∣ u ∣]_{1}^{1 + x}

Simplify

x \frac{1}{x} [ln ∣ u ∣]_{1}^{1 + x} : [ln ∣ u ∣]_{1}^{1 + x}

= [ln ∣ u ∣]_{1}^{1 + x}

Compute the boundaries:

ln ∣ x + 1 ∣

= ln ∣ x + 1 ∣

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What is the integral from 0 to 1 of x/(1+xy) ?
The integral from 0 to 1 of x/(1+xy) is ln|x+1|