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

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

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integral of (y^2-1)^2

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

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

Solution steps

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

Expand

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

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

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

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

\int 1 d y = y

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} + a^{2})

\int e^{3 x - 4} d x

\frac{\partial}{\partial x} (e^{x e^{y}})

f^{^{'}} (x) = 7 e^{- e} - 5 e^{- 7 x}

x \to 0 + lim (\frac{3}{1 + e ^{\frac{1}{x}}})

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