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

The integral of (2x^2)/(3-4x^3) is -1/6 ln|3-4x^3|+C

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

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

- \frac{1}{6} ln 3 - 4 x^{3} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{2}}{3 - 4 x ^{3}} d x

Apply u-substitution

: \int - \frac{1}{12 u} d u

= 2 \cdot \int - \frac{1}{12 u} d u

Take the constant out:

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

= 2 (- \frac{1}{12} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{12} ln ∣ u ∣)

Substitute back

u = 3 - 4 x^{3}

= 2 (- \frac{1}{12} ln 3 - 4 x^{3})

Simplify

2 (- \frac{1}{12} ln 3 - 4 x^{3}) : - \frac{1}{6} ln 3 - 4 x^{3}

= - \frac{1}{6} ln 3 - 4 x^{3}

Add a constant to the solution

= - \frac{1}{6} ln 3 - 4 x^{3} + C

- 2

\frac{d}{d x} (arctan (x y))

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

f (x) = \frac{3}{1 + x ^{2}}

\int_{0}^{1} 3 x^{2} (x^{3} + 5)^{- 2} d x

What is the integral of (2x^2)/(3-4x^3) ?
The integral of (2x^2)/(3-4x^3) is -1/6 ln|3-4x^3|+C