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

The integral of (4-x^2)^2x^2 is (16x^3)/3-(8x^5)/5+(x^7)/7+C

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

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

\frac{16 x ^{3}}{3} - \frac{8 x ^{5}}{5} + \frac{x ^{7}}{7} + C

Solution steps

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

Expand

(4 - x^{2})^{2} x^{2} : 16 x^{2} - 8 x^{4} + x^{6}

= \int 16 x^{2} - 8 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 16 x^{2} d x - \int 8 x^{4} d x + \int x^{6} d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 8 x^{4} d x = \frac{8 x ^{5}}{5}

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

= \frac{16 x ^{3}}{3} - \frac{8 x ^{5}}{5} + \frac{x ^{7}}{7}

Add a constant to the solution

= \frac{16 x ^{3}}{3} - \frac{8 x ^{5}}{5} + \frac{x ^{7}}{7} + C

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

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

e^{s^{2} + 1}

\frac{d y}{d x} - \frac{1}{x} y = 1 - \frac{1}{x}

f (x) = cos (6 x)

What is the integral of (4-x^2)^2x^2 ?
The integral of (4-x^2)^2x^2 is (16x^3)/3-(8x^5)/5+(x^7)/7+C