What is the integral from 0 to 2 of (6x^2)/(2+x^3) ?

The integral from 0 to 2 of (6x^2)/(2+x^3) is 2ln(5)

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integral from 0 to 2 of (6x^2)/(2+x^3)

\int_{0}^{2} \frac{6 x ^{2}}{2 + x ^{3}} d x

2 ln (5)

Decimal Notation

3.21887 \dots

Solution steps

\int_{0}^{2} \frac{6 x ^{2}}{2 + x ^{3}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} \frac{x ^{2}}{2 + x ^{3}} d x

Apply u-substitution

: \int_{2}^{10} \frac{1}{3 u} d u

= 6 \cdot \int_{2}^{10} \frac{1}{3 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int_{2}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{10}

Simplify

6 \cdot \frac{1}{3} [ln ∣ u ∣]_{2}^{10} : 2 [ln ∣ u ∣]_{2}^{10}

= 2 [ln ∣ u ∣]_{2}^{10}

Compute the boundaries:

ln (10) - ln (2)

= 2 (ln (10) - ln (2))

Simplify

= 2 ln (5)

\int_{1}^{5} ln (2 x) - ln (x) d x

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\int_{- 1}^{1} x^{2} - 2 x d x

\int_{0}^{2 π} 2 cos^{2} (t) d t

\int_{1}^{4} \frac{ln ( x )}{x} d x

What is the integral from 0 to 2 of (6x^2)/(2+x^3) ?
The integral from 0 to 2 of (6x^2)/(2+x^3) is 2ln(5)