What is the integral from 0 to 1 of (16x)/(x^2+1) ?

The integral from 0 to 1 of (16x)/(x^2+1) is 8ln(2)

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integral from 0 to 1 of (16x)/(x^2+1)

\int_{0}^{1} \frac{16 x}{x ^{2} + 1} d x

8 ln (2)

Decimal Notation

5.54517 \dots

Solution steps

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

Apply u-substitution

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

= 16 \cdot \int_{1}^{2} \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_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= 8 [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= 8 ln (2)

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What is the integral from 0 to 1 of (16x)/(x^2+1) ?
The integral from 0 to 1 of (16x)/(x^2+1) is 8ln(2)