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

The integral from 1 to 8 of (10x)/(x^2+6) is 5ln(10)

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integral from 1 to 8 of (10x)/(x^2+6)

\int_{1}^{8} \frac{10 x}{x ^{2} + 6} d x

5 ln (10)

Decimal

11.51292 \dots

Solution steps

\int_{1}^{8} \frac{10 x}{x ^{2} + 6} d x

Take the constant out:

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

= 10 \cdot \int_{1}^{8} \frac{x}{x ^{2} + 6} d x

Apply u-substitution

: \int_{7}^{70} \frac{1}{2 u} d u

= 10 \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

= 10 \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 ∣)

= 10 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70}

Simplify

10 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{70} : 5 [ln ∣ u ∣]_{7}^{70}

= 5 [ln ∣ u ∣]_{7}^{70}

Compute the boundaries:

ln (70) - ln (7)

= 5 (ln (70) - ln (7))

Simplify

= 5 ln (10)

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What is the integral from 1 to 8 of (10x)/(x^2+6) ?
The integral from 1 to 8 of (10x)/(x^2+6) is 5ln(10)