What is the integral from 1 to 3 of x/(x^2+4) ?

The integral from 1 to 3 of x/(x^2+4) is 1/2 (ln(13)-ln(5))

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integral from 1 to 3 of x/(x^2+4)

\int_{1}^{3} \frac{x}{x ^{2} + 4} d x

\frac{1}{2} (ln (13) - ln (5))

Solution steps

\int_{1}^{3} \frac{x}{x ^{2} + 4} d x

Apply u-substitution

: \int_{5}^{13} \frac{1}{2 u} d u

= \int_{5}^{13} \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_{5}^{13} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{13}

Compute the boundaries:

ln (13) - ln (5)

= \frac{1}{2} (ln (13) - ln (5))

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What is the integral from 1 to 3 of x/(x^2+4) ?
The integral from 1 to 3 of x/(x^2+4) is 1/2 (ln(13)-ln(5))