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

The integral from 0 to 2 of (x^2)/(x^3+3) is 1/3 (ln(11)-ln(3))

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

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

\frac{1}{3} (ln (11) - ln (3))

Solution steps

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

Apply u-substitution

: \int_{3}^{11} \frac{1}{3 u} d u

= \int_{3}^{11} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{3}^{11} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{3}^{11}

Compute the boundaries:

ln (11) - ln (3)

= \frac{1}{3} (ln (11) - ln (3))

\int_{0}^{1} x sinh (x) d x

\int_{0}^{3} x^{2} - 3 d x

\int_{0}^{\frac{π}{4}} tan (θ) d θ

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (π \cdot x) d x

\int_{0}^{3} x^{3} - 6 x^{2} + 8 x d x

What is the integral from 0 to 2 of (x^2)/(x^3+3) ?
The integral from 0 to 2 of (x^2)/(x^3+3) is 1/3 (ln(11)-ln(3))