What is the integral of (-x^2+6)^2-(x^2+4x)^2 ?

The integral of (-x^2+6)^2-(x^2+4x)^2 is -2x^4-4x^3-(16x^3)/3+36x+C

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integral of (-x^2+6)^2-(x^2+4x)^2

\int (- x^{2} + 6)^{2} - (x^{2} + 4 x)^{2} d x

- 2 x^{4} - 4 x^{3} - \frac{16 x ^{3}}{3} + 36 x + C

Solution steps

\int (- x^{2} + 6)^{2} - (x^{2} + 4 x)^{2} 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^{2} + 6)^{2} d x - \int (x^{2} + 4 x)^{2} d x

\int (- x^{2} + 6)^{2} d x = \frac{x ^{5}}{5} - 4 x^{3} + 36 x

\int (x^{2} + 4 x)^{2} d x = \frac{x ^{5}}{5} + 2 x^{4} + \frac{16 x ^{3}}{3}

= \frac{x ^{5}}{5} - 4 x^{3} + 36 x - (\frac{x ^{5}}{5} + 2 x^{4} + \frac{16 x ^{3}}{3})

Simplify

= - 2 x^{4} - 4 x^{3} - \frac{16 x ^{3}}{3} + 36 x

Add a constant to the solution

= - 2 x^{4} - 4 x^{3} - \frac{16 x ^{3}}{3} + 36 x + C

2 x^{3}

- 27

\frac{1}{s ( s ^{2} + 16 )}

\frac{d}{d s} (\frac{1}{s ^{2} + 1})

4 x^{3} e^{- x}

What is the integral of (-x^2+6)^2-(x^2+4x)^2 ?
The integral of (-x^2+6)^2-(x^2+4x)^2 is -2x^4-4x^3-(16x^3)/3+36x+C