What is the integral of (x^2)/(3x^3+4) ?

The integral of (x^2)/(3x^3+4) is 1/9 ln|3x^3+4|+C

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integral of (x^2)/(3x^3+4)

\int \frac{x ^{2}}{3 x ^{3} + 4} d x

\frac{1}{9} ln 3 x^{3} + 4 + C

Solution steps

\int \frac{x ^{2}}{3 x ^{3} + 4} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 3 x^{3} + 4

= \frac{1}{9} ln 3 x^{3} + 4

Add a constant to the solution

= \frac{1}{9} ln 3 x^{3} + 4 + C

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What is the integral of (x^2)/(3x^3+4) ?
The integral of (x^2)/(3x^3+4) is 1/9 ln|3x^3+4|+C