What is the integral of (y^3)/((2+3y^4)^3) ?

The integral of (y^3)/((2+3y^4)^3) is -1/(24(2+3y^4)^2)+C

integral of (y^3)/((2+3y^4)^3)

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

- \frac{1}{24 ( 2 + 3 y ^{4} ) ^{2}} + C

Solution steps

\int \frac{y ^{3}}{( 2 + 3 y ^{4} ) ^{3}} d y

Apply u-substitution

= \int \frac{1}{12 u ^{3}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= \frac{1}{12} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 + 3 y^{4}

= \frac{1}{12} (- \frac{1}{2 ( 2 + 3 y ^{4} ) ^{2}})

Simplify

\frac{1}{12} (- \frac{1}{2 ( 2 + 3 y ^{4} ) ^{2}}) : - \frac{1}{24 ( 2 + 3 y ^{4} ) ^{2}}

= - \frac{1}{24 ( 2 + 3 y ^{4} ) ^{2}}

Add a constant to the solution

= - \frac{1}{24 ( 2 + 3 y ^{4} ) ^{2}} + C

sin (tan (3 x))

lo g_{4} (e^{2 x} cos (3 x))

\frac{30 x ^{2}}{( - 5 x ^{3} + 4 ) ^{3}}

x^{2} + x y - y^{2} = - 11, (1, 4)

t^{2} y^{^{''}} + 3 t y^{^{'}} - 3 y = \frac{1}{t}

What is the integral of (y^3)/((2+3y^4)^3) ?
The integral of (y^3)/((2+3y^4)^3) is -1/(24(2+3y^4)^2)+C