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

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

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integral of (4-x^2)^2

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

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

Solution steps

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

Expand

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

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

\int 16 d x = 16 x

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

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

Add a constant to the solution

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

\int_{2}^{4} \frac{6 + u ^{2}}{u ^{3}} d u

x \to 4 lim (2 x + 1)

\frac{d}{d y} (e^{y^{2}})

4 - x^{2}, 0, 2

x \to \frac{π}{2} lim (sin (2 x))

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