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Values: \(n = 3, 6, 9, 12, 15, 18\) give \(n^3 = 27, 216, 729, 1728, 3375, 5832\) — all less than 9999. That is 6 values.
If $n = 2^3$, what is the value of $n^n$?
$n = 2^3 = 8$.
$n^n = 8^8 = (2^3)^8 = 2^{3 \times 8} = 2^{24}$.
Answer: $\mathbf{2^{24}}$.
The integral $\displaystyle\int\dfrac{2x^{12}+5x^9}{(x^5+x^3+1)^3}\,dx$ equals (where $C$ is an arbitrary constant):
Divide numerator and denominator by $x^{15}$:
$\displaystyle\int\dfrac{2x^{-3}+5x^{-6}}{(1+x^{-2}+x^{-5})^3}\,dx$
Let $t=1+x^{-2}+x^{-5}$. Then $dt=(-2x^{-3}-5x^{-6})dx$, so the numerator (divided by $x^{15}$) gives $-dt$:
$=-\displaystyle\int\dfrac{dt}{t^3} \cdot \dfrac{1}{x^{15}/x^{15}}$
Wait — let me redo with $t = x^{-5}+x^{-3}+1$... Actually substituting directly: $t=1+x^{-2}+x^{-5}$, $-dt=(2x^{-3}+5x^{-6})dx$.
Integral $= -\int t^{-3}dt = \dfrac{1}{2t^2}+C = \dfrac{x^{10}}{2(x^5+x^3+1)^2}+C$
If $y = \sec(\tan^{-1}x)$, find $\dfrac{dy}{dx}$ at $x=1$.
Let $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $y = \sec\theta$.
$\frac{dy}{dx} = \sec\theta\tan\theta\cdot\frac{d\theta}{dx} = \sec\theta\tan\theta\cdot\frac{1}{1+x^2}$
Since $\tan\theta=x$ and $\sec\theta=\sqrt{1+x^2}$:
$\frac{dy}{dx} = \sqrt{1+x^2}\cdot x\cdot\frac{1}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$
At $x=1$: $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$
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