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

The integral from 0 to 1 of (8t)/(t^2+1) is 4ln(2)

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

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

4 ln (2)

Decimal

2.77258 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 8 \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

= 8 \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 ∣)

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

Simplify

8 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{2} : 4 [ln ∣ u ∣]_{1}^{2}

= 4 [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= 4 ln (2)

What is the integral from 0 to 1 of (8t)/(t^2+1) ?
The integral from 0 to 1 of (8t)/(t^2+1) is 4ln(2)