What is the integral from 1 to 8 of (16x)/(x^2+6) ?

The integral from 1 to 8 of (16x)/(x^2+6) is 8ln(10)

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integral from 1 to 8 of (16x)/(x^2+6)

\int_{1}^{8} \frac{16 x}{x ^{2} + 6} d x

8 ln (10)

Decimal

18.42068 \dots

Solution steps

\int_{1}^{8} \frac{16 x}{x ^{2} + 6} d x

Take the constant out:

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

= 16 \cdot \int_{1}^{8} \frac{x}{x ^{2} + 6} d x

Apply u-substitution

= 16 \cdot \int_{7}^{70} \frac{1}{2 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{2} \cdot \int_{7}^{70} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70}

Simplify

16 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70} : 8 [ln ∣ u ∣]_{7}^{70}

= 8 [ln ∣ u ∣]_{7}^{70}

Compute the boundaries:

ln (70) - ln (7)

= 8 (ln (70) - ln (7))

Simplify

= 8 ln (10)

What is the integral from 1 to 8 of (16x)/(x^2+6) ?
The integral from 1 to 8 of (16x)/(x^2+6) is 8ln(10)