What is the integral from 0 to 5 of x/(x^2+5) ?

The integral from 0 to 5 of x/(x^2+5) is 1/2 ln(6)

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integral from 0 to 5 of x/(x^2+5)

\int_{0}^{5} \frac{x}{x ^{2} + 5} d x

\frac{1}{2} ln (6)

Decimal Notation

0.89587 \dots

Solution steps

\int_{0}^{5} \frac{x}{x ^{2} + 5} d x

Apply u-substitution

: \int_{5}^{30} \frac{1}{2 u} d u

= \int_{5}^{30} \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}^{30} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{30}

Compute the boundaries:

ln (30) - ln (5)

= \frac{1}{2} (ln (30) - ln (5))

Simplify

= \frac{1}{2} ln (6)

\int_{2}^{5} (x^{3} + x - 4) d x

\int_{0}^{1} (a + x)^{2} - x^{2} d a

\int_{0}^{- 1} (x ln (x)) d x

\int_{c}^{d} x d x

\int_{1}^{4} \frac{1}{6} d x

What is the integral from 0 to 5 of x/(x^2+5) ?
The integral from 0 to 5 of x/(x^2+5) is 1/2 ln(6)