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

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

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

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

ln (2)

Solution steps

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

Apply u-substitution

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

= 2 \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

= 2 \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 ∣)

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

Simplify

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

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

y = sin (4 x)

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