What is the integral from 2 to 7 of (16x)/(x^2+1) ?

The integral from 2 to 7 of (16x)/(x^2+1) is 8ln(10)

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integral from 2 to 7 of (16x)/(x^2+1)

\int_{2}^{7} \frac{16 x}{x ^{2} + 1} d x

8 ln (10)

Decimal Notation

18.42068 \dots

Solution steps

\int_{2}^{7} \frac{16 x}{x ^{2} + 1} d x

Take the constant out:

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

= 16 \cdot \int_{2}^{7} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

= 16 \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

= 16 \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 ∣)

= 16 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50}

Simplify

16 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{50} : 8 [ln ∣ u ∣]_{5}^{50}

= 8 [ln ∣ u ∣]_{5}^{50}

Compute the boundaries:

ln (50) - ln (5)

= 8 (ln (50) - ln (5))

Simplify

= 8 ln (10)

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What is the integral from 2 to 7 of (16x)/(x^2+1) ?
The integral from 2 to 7 of (16x)/(x^2+1) is 8ln(10)