What is the integral of (1-3y)^2-(1-y^2)^2 ?

The integral of (1-3y)^2-(1-y^2)^2 is -(y^5)/5+(11y^3)/3-3y^2+C

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integral of (1-3y)^2-(1-y^2)^2

\int (1 - 3 y)^{2} - (1 - y^{2})^{2} d y

- \frac{y ^{5}}{5} + \frac{11 y ^{3}}{3} - 3 y^{2} + C

Solution steps

\int (1 - 3 y)^{2} - (1 - y^{2})^{2} d y

Simplify

(1 - 3 y)^{2} - (1 - y^{2})^{2} : - y^{4} + 11 y^{2} - 6 y

= \int - y^{4} + 11 y^{2} - 6 y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int y^{4} d y + \int 11 y^{2} d y - \int 6 y d y

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

\int 11 y^{2} d y = \frac{11 y ^{3}}{3}

\int 6 y d y = 3 y^{2}

= - \frac{y ^{5}}{5} + \frac{11 y ^{3}}{3} - 3 y^{2}

Add a constant to the solution

= - \frac{y ^{5}}{5} + \frac{11 y ^{3}}{3} - 3 y^{2} + C

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What is the integral of (1-3y)^2-(1-y^2)^2 ?
The integral of (1-3y)^2-(1-y^2)^2 is -(y^5)/5+(11y^3)/3-3y^2+C