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

The integral of x^4(1-x^2) is (x^5)/5-(x^7)/7+C

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integral of x^4(1-x^2)

\int x^{4} (1 - x^{2}) d x

\frac{x ^{5}}{5} - \frac{x ^{7}}{7} + C

Solution steps

\int x^{4} (1 - x^{2}) d x

Expand

x^{4} (1 - x^{2}) : x^{4} - x^{6}

= \int x^{4} - x^{6} 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 x^{4} d x - \int x^{6} d x

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

\int x^{6} d x = \frac{x ^{7}}{7}

= \frac{x ^{5}}{5} - \frac{x ^{7}}{7}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + C

f (x) = \frac{1}{( 1 + x )} 1

x^{^{'}} - a x - 1 = 0

\int_{0}^{1} x \cdot 3^{x^{2} + 1} d x

y^{^{''}} + 9 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (x^{3} + x e^{y} - 2 y)

What is the integral of x^4(1-x^2) ?
The integral of x^4(1-x^2) is (x^5)/5-(x^7)/7+C