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

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

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

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

ln (2)

Decimal

0.69314 \dots

Solution steps

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

Take the constant out:

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

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

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)

x^{2} y^{^{''}} - 3 x y^{^{'}} + 5 y = 0

f (x) = 2 x^{2} - 9 x + 10, at x = 2

\frac{d}{d x} (x arcsin (x))

tan (1 + e^{2 t})

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 9))

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