What is the integral from 0 to 1 of ((3t^2)/(1+t^3)) ?

The integral from 0 to 1 of ((3t^2)/(1+t^3)) is ln(2)

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integral from 0 to 1 of ((3t^2)/(1+t^3))

\int_{0}^{1} (\frac{3 t ^{2}}{1 + t ^{3}}) d t

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{1} \frac{3 t ^{2}}{1 + t ^{3}} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{t ^{2}}{1 + t ^{3}} d t

Apply u-substitution

: \int_{1}^{2} \frac{1}{3 u} d u

= 3 \cdot \int_{1}^{2} \frac{1}{3 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{2}

Simplify

3 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{2} : [ln ∣ u ∣]_{1}^{2}

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

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What is the integral from 0 to 1 of ((3t^2)/(1+t^3)) ?
The integral from 0 to 1 of ((3t^2)/(1+t^3)) is ln(2)