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

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

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integral of (1-x^2)^2

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

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

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

\int 1 d x = x

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

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

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

Add a constant to the solution

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

3 6 x^{4} + 7 x^{3}

\frac{d}{d x} ((6 x + 4) (3 x^{2} + 2))

x \to - \infty lim (\frac{5 x ^{4} + e ^{5 x}}{x ^{4}})

\frac{d ^{2}}{d x ^{2}} ((5 + \frac{2}{x})^{3})

(4 x + 4 y)^{3} = 64 x^{3} + 64 y^{3}

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