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

The integral of ((y^2)/4)^2 is (y^5)/(16)-(y^5)/(20)+C

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

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

\frac{y ^{5}}{16} - \frac{y ^{5}}{20} + C

Solution steps

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

Apply Integration By Parts

: \frac{y ^{5}}{16} - \int \frac{y ^{4}}{4} d y

= \frac{y ^{5}}{16} - \int \frac{y ^{4}}{4} d y

\int \frac{y ^{4}}{4} d y = \frac{y ^{5}}{20}

= \frac{y ^{5}}{16} - \frac{y ^{5}}{20}

Add a constant to the solution

= \frac{y ^{5}}{16} - \frac{y ^{5}}{20} + C

\frac{d}{d x} (e^{x^{3} + 6 x^{2}})

\int \frac{1}{10 ( x + 2 )} d x

\int 4 cot^{3} (x) csc^{3} (x) d x

y = 5 2 x^{3} + 8 x, (2, 2)

x \to 0 lim (\frac{sin ( 0 )}{0})

What is the integral of ((y^2)/4)^2 ?
The integral of ((y^2)/4)^2 is (y^5)/(16)-(y^5)/(20)+C