What is the integral of (3x^3)/(sqrt(2-9x^4)) ?

The integral of (3x^3)/(sqrt(2-9x^4)) is -1/6 sqrt(2-9x^4)+C

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integral of (3x^3)/(sqrt(2-9x^4))

\int \frac{3 x ^{3}}{2 - 9 x ^{4}} d x

- \frac{1}{6} 2 - 9 x^{4} + C

Solution steps

\int \frac{3 x ^{3}}{2 - 9 x ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{3}}{2 - 9 x ^{4}} d x

Apply u-substitution

= 3 \cdot \int - \frac{1}{36 u} d u

Take the constant out:

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

= 3 (- \frac{1}{36} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

= 3 (- \frac{1}{36} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 2 - 9 x^{4}

= 3 (- \frac{1}{36} \cdot 2 (2 - 9 x^{4})^{\frac{1}{2}})

Simplify

3 (- \frac{1}{36} \cdot 2 (2 - 9 x^{4})^{\frac{1}{2}}) : - \frac{1}{6} 2 - 9 x^{4}

= - \frac{1}{6} 2 - 9 x^{4}

Add a constant to the solution

= - \frac{1}{6} 2 - 9 x^{4} + C

What is the integral of (3x^3)/(sqrt(2-9x^4)) ?
The integral of (3x^3)/(sqrt(2-9x^4)) is -1/6 sqrt(2-9x^4)+C