What is the integral of (5x^2)/(3-3x^3) ?

The integral of (5x^2)/(3-3x^3) is -5/9 ln|3-3x^3|+C

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integral of (5x^2)/(3-3x^3)

\int \frac{5 x ^{2}}{3 - 3 x ^{3}} d x

- \frac{5}{9} ln 3 - 3 x^{3} + C

Solution steps

\int \frac{5 x ^{2}}{3 - 3 x ^{3}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{3 - 3 x ^{3}} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 (- \frac{1}{9} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 (- \frac{1}{9} ln ∣ u ∣)

Substitute back

u = 3 - 3 x^{3}

= 5 (- \frac{1}{9} ln 3 - 3 x^{3})

Simplify

5 (- \frac{1}{9} ln 3 - 3 x^{3}) : - \frac{5}{9} ln 3 - 3 x^{3}

= - \frac{5}{9} ln 3 - 3 x^{3}

Add a constant to the solution

= - \frac{5}{9} ln 3 - 3 x^{3} + C

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What is the integral of (5x^2)/(3-3x^3) ?
The integral of (5x^2)/(3-3x^3) is -5/9 ln|3-3x^3|+C