What is the integral of (y^2)/(4-1) ?

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

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integral of (y^2)/(4-1)

integral

\frac{y ^{2}}{4 - 1}

\frac{y ^{3}}{9} + C

Solution steps

\int \frac{y ^{2}}{4 - 1} d y

Take the constant out:

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

= \frac{1}{4 - 1} \cdot \int y^{2} d y

Simplify

= \frac{1}{3} \cdot \int y^{2} d y

Apply the Power Rule

: \frac{y ^{3}}{3}

= \frac{1}{3} \cdot \frac{y ^{3}}{3}

Simplify

\frac{1}{3} \cdot \frac{y ^{3}}{3} : \frac{y ^{3}}{9}

= \frac{y ^{3}}{9}

Add a constant to the solution

= \frac{y ^{3}}{9} + C

(1 - n) cos (x)

\frac{1}{( r - 1 )}

\frac{cos ( 1 )}{x}

\frac{1}{( sin ^{2} ( x ) \cdot cos ^{4} ( x ) )}

\frac{x ^{4}}{( x + 1 )}