What is the integral of (2x^5-3x^2)/(1+3x^3-x^6) ?

The integral of (2x^5-3x^2)/(1+3x^3-x^6) is -1/3 ln|1+3x^3-x^6|+C

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integral of (2x^5-3x^2)/(1+3x^3-x^6)

\int \frac{2 x ^{5} - 3 x ^{2}}{1 + 3 x ^{3} - x ^{6}} d x

- \frac{1}{3} ln 1 + 3 x^{3} - x^{6} + C

Solution steps

\int \frac{2 x ^{5} - 3 x ^{2}}{1 + 3 x ^{3} - x ^{6}} 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 = 1 + 3 x^{3} - x^{6}

= - \frac{1}{3} ln 1 + 3 x^{3} - x^{6}

Add a constant to the solution

= - \frac{1}{3} ln 1 + 3 x^{3} - x^{6} + C

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What is the integral of (2x^5-3x^2)/(1+3x^3-x^6) ?
The integral of (2x^5-3x^2)/(1+3x^3-x^6) is -1/3 ln|1+3x^3-x^6|+C