What is the integral of (3x^3)/(sqrt(5-9x^4)) ?

The integral of (3x^3)/(sqrt(5-9x^4)) is -1/6 sqrt(5-9x^4)+C

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integral of (3x^3)/(sqrt(5-9x^4))

\int \frac{3 x ^{3}}{5 - 9 x ^{4}} d x

- \frac{1}{6} 5 - 9 x^{4} + C

Solution steps

\int \frac{3 x ^{3}}{5 - 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}}{5 - 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 = 5 - 9 x^{4}

= 3 (- \frac{1}{36} \cdot 2 (5 - 9 x^{4})^{\frac{1}{2}})

Simplify

3 (- \frac{1}{36} \cdot 2 (5 - 9 x^{4})^{\frac{1}{2}}) : - \frac{1}{6} 5 - 9 x^{4}

= - \frac{1}{6} 5 - 9 x^{4}

Add a constant to the solution

= - \frac{1}{6} 5 - 9 x^{4} + C

\int \frac{x ^{2} + 6 x + 2}{x ^{3} + 9 x ^{2} + 6 x + 3} d x

t an g e n t - 3 cos (x), at x = - \frac{π}{3}

ma c l a u r in e^{2 x^{3}}

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 1)

l a pl a ce t r an s f or m cos^{2} (13 t)

What is the integral of (3x^3)/(sqrt(5-9x^4)) ?
The integral of (3x^3)/(sqrt(5-9x^4)) is -1/6 sqrt(5-9x^4)+C