What is the integral from 0 to 1 of (12x)/(x^2+7) ?

The integral from 0 to 1 of (12x)/(x^2+7) is 6(3ln(2)-ln(7))

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integral from 0 to 1 of (12x)/(x^2+7)

\int_{0}^{1} \frac{12 x}{x ^{2} + 7} d x

6 (3 ln (2) - ln (7))

Decimal Notation

0.80118 \dots

Solution steps

\int_{0}^{1} \frac{12 x}{x ^{2} + 7} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{1} \frac{x}{x ^{2} + 7} d x

Apply u-substitution

: \int_{7}^{8} \frac{1}{2 u} d u

= 12 \cdot \int_{7}^{8} \frac{1}{2 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{2} \cdot \int_{7}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 12 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{8}

Simplify

12 \cdot \frac{1}{2} [ln ∣ u ∣]_{7}^{8} : 6 [ln ∣ u ∣]_{7}^{8}

= 6 [ln ∣ u ∣]_{7}^{8}

Compute the boundaries:

3 ln (2) - ln (7)

= 6 (3 ln (2) - ln (7))

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What is the integral from 0 to 1 of (12x)/(x^2+7) ?
The integral from 0 to 1 of (12x)/(x^2+7) is 6(3ln(2)-ln(7))