What is the integral from 0 to 8 of (2x)/(x^2+1) ?

The integral from 0 to 8 of (2x)/(x^2+1) is ln(65)

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integral from 0 to 8 of (2x)/(x^2+1)

\int_{0}^{8} \frac{2 x}{x ^{2} + 1} d x

ln (65)

Decimal

4.17438 \dots

Solution steps

\int_{0}^{8} \frac{2 x}{x ^{2} + 1} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{8} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{1}^{65} \frac{1}{2 u} d u

= 2 \cdot \int_{1}^{65} \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{65} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{65}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{65} : [ln ∣ u ∣]_{1}^{65}

= [ln ∣ u ∣]_{1}^{65}

Compute the boundaries:

ln (65)

= ln (65)

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What is the integral from 0 to 8 of (2x)/(x^2+1) ?
The integral from 0 to 8 of (2x)/(x^2+1) is ln(65)