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

The integral from 0 to 2 of (2t)/(1+t^2) is ln(5)

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

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

ln (5)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int_{1}^{5} \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}^{5} \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}^{5}

Simplify

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

= [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= ln (5)

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\int_{- 2}^{2} (x^{2} - 5) d x

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What is the integral from 0 to 2 of (2t)/(1+t^2) ?
The integral from 0 to 2 of (2t)/(1+t^2) is ln(5)