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

The integral of (x^2)/(3+x^3) is 1/3 ln|3+x^3|+C

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integral of (x^2)/(3+x^3)

\int \frac{x ^{2}}{3 + x ^{3}} d x

\frac{1}{3} ln 3 + x^{3} + C

Solution steps

\int \frac{x ^{2}}{3 + x ^{3}} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 + x^{3}

= \frac{1}{3} ln 3 + x^{3}

Add a constant to the solution

= \frac{1}{3} ln 3 + x^{3} + C

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What is the integral of (x^2)/(3+x^3) ?
The integral of (x^2)/(3+x^3) is 1/3 ln|3+x^3|+C