What is the integral from 2 to 7 of (6x)/(x^2+1) ?

The integral from 2 to 7 of (6x)/(x^2+1) is 3ln(10)

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integral from 2 to 7 of (6x)/(x^2+1)

\int_{2}^{7} \frac{6 x}{x ^{2} + 1} d x

3 ln (10)

Decimal Notation

6.90775 \dots

Solution steps

\int_{2}^{7} \frac{6 x}{x ^{2} + 1} d x

Take the constant out:

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

= 6 \cdot \int_{2}^{7} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{5}^{50} \frac{1}{2 u} d u

= 6 \cdot \int_{5}^{50} \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{5}^{50} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50}

Simplify

6 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50} : 3 [ln ∣ u ∣]_{5}^{50}

= 3 [ln ∣ u ∣]_{5}^{50}

Compute the boundaries:

ln (50) - ln (5)

= 3 (ln (50) - ln (5))

Simplify

= 3 ln (10)

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\int_{2}^{4} (- x^{2} + 6 x - 8)^{2} d x

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What is the integral from 2 to 7 of (6x)/(x^2+1) ?
The integral from 2 to 7 of (6x)/(x^2+1) is 3ln(10)