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

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

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

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

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

Solution steps

\int \frac{x ^{2}}{x ^{3} - 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 = x^{3} - 3

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

Add a constant to the solution

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

\int_{0}^{5} 0.2 x d x

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x \to 3 lim (x^{2} - 3 x)

\int (a^{\frac{2}{3}} - x^{\frac{2}{3}})^{2} d x

\int cos^{2} (6 x) sin^{3} (6 x) d x

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