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

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

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

\int_{0}^{e} \frac{x}{x ^{2} + e} d x

\frac{1}{2} (ln (e^{2} + e) - 1)

Decimal Notation

0.65663 \dots

Solution steps

\int_{0}^{e} \frac{x}{x ^{2} + e} d x

Apply u-substitution

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

= \int_{e}^{e^{2} + e} \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_{e}^{e^{2} + e} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{e}^{e^{2} + e}

Compute the boundaries:

ln (e^{2} + e) - 1

= \frac{1}{2} (ln (e^{2} + e) - 1)

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