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

The integral of (2x^2)/(1-6x^3) is -1/9 ln|1-6x^3|+C

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

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

- \frac{1}{9} ln 1 - 6 x^{3} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{18 u} d u

= 2 \cdot \int - \frac{1}{18 u} d u

Take the constant out:

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

= 2 (- \frac{1}{18} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{18} ln ∣ u ∣)

Substitute back

u = 1 - 6 x^{3}

= 2 (- \frac{1}{18} ln 1 - 6 x^{3})

Simplify

2 (- \frac{1}{18} ln 1 - 6 x^{3}) : - \frac{1}{9} ln 1 - 6 x^{3}

= - \frac{1}{9} ln 1 - 6 x^{3}

Add a constant to the solution

= - \frac{1}{9} ln 1 - 6 x^{3} + C

\frac{\partial}{\partial x} (sin (y) - y sin (x))

f (x) = 3 x^{3} - x^{2} + 2, (1, 4)

\int t^{3} \cdot ln (t^{2}) d t

\frac{\partial}{\partial y} (x y^{3} - x^{2})

\frac{\partial}{\partial y} (x e^{- x^{2} y})

What is the integral of (2x^2)/(1-6x^3) ?
The integral of (2x^2)/(1-6x^3) is -1/9 ln|1-6x^3|+C