What is the integral from 4 to 6 of x/(4x^2+8) ?

The integral from 4 to 6 of x/(4x^2+8) is 1/8 ln(19/9)

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integral from 4 to 6 of x/(4x^2+8)

\int_{4}^{6} \frac{x}{4 x ^{2} + 8} d x

\frac{1}{8} ln (\frac{19}{9})

Solution steps

\int_{4}^{6} \frac{x}{4 x ^{2} + 8} d x

Apply u-substitution

: \int_{72}^{152} \frac{1}{8 u} d u

= \int_{72}^{152} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{72}^{152} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} [ln ∣ u ∣]_{72}^{152}

Compute the boundaries:

ln (152) - ln (72)

= \frac{1}{8} (ln (152) - ln (72))

Simplify

= \frac{1}{8} ln (\frac{19}{9})

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What is the integral from 4 to 6 of x/(4x^2+8) ?
The integral from 4 to 6 of x/(4x^2+8) is 1/8 ln(19/9)