Determine the following integrals: 1 integral (1)/(-2x2+20x-47) dx 2 integral 3x 2x dx 3 integral e⁻²x [3]x6 dx 4 integral 5 4x dx 5 integral ^-1((7x)/(3)) dx

$$-2x^2 + 20x - 47 = -2(x^2 - 10x + \frac{47}{2})$$ $$= -2((x - 5)^2 - 25 + \frac{47}{2})$$ $$= -2((x - 5)^2 - \frac{50}{2} + \frac{47}{2})$$ $$= -2((x - 5)^2 - \frac{3}{2})$$ $$= 3 - 2(x - 5)^2$$

Let $u = x - 5$, so $du = dx$.
The integral becomes:

$$\int \frac{1}{3 - 2u^2} du = \frac{1}{2} \int \frac{1}{\frac{3}{2} - u^2} du$$

Here, $a^2 = \frac{3}{2}$, so $a = \sqrt{\frac{3}{2}}$.

$$\frac{1}{2} \cdot \frac{1}{2\sqrt{\frac{3}{2}}} \ln\left|\frac{\sqrt{\frac{3}{2}} + u}{\sqrt{\frac{3}{2}} - u}\right| + C$$ $$= \frac{1}{4\frac{\sqrt{3}}{\sqrt{2}}} \ln\left|\frac{\frac{\sqrt{3}}{\sqrt{2}} + u}{\frac{\sqrt{3}}{\sqrt{2}} - u}\right| + C$$ $$= \frac{\sqrt{2}}{4\sqrt{3}} \ln\left|\frac{\sqrt{3} + \sqrt{2}u}{\sqrt{3} - \sqrt{2}u}\right| + C$$

Rationalize the coefficient: $\frac{\sqrt{2}}{4\sqrt{3}} = \frac{\sqrt{2}\sqrt{3}}{4\sqrt{3}\sqrt{3}} = \frac{\sqrt{6}}{12}$.

Let $A = 3x$ and $B = 2x$.

$$\sin(3x)\cos(2x) = \frac{1}{2}[\sin(3x+2x) + \sin(3x-2x)]$$ $$= \frac{1}{2}[\sin(5x) + \sin(x)]$$

$$\int \frac{1}{2}[\sin(5x) + \sin(x)] dx = \frac{1}{2} \int (\sin(5x) + \sin(x)) dx$$ $$= \frac{1}{2} \left( -\frac{1}{5}\cos(5x) - \cos(x) \right) + C$$ $$\boxed{-\frac{1}{10}\cos(5x) - \frac{1}{2}\cos(x) + C}$$

$$\int x^2 e^{-2x} dx$$

This integral requires integration by parts.

Let $u = x^2 \implies du = 2x \, dx$.
Let $dv = e^{-2x} dx \implies v = \int e^{-2x} dx = -\frac{1}{2}e^{-2x}$.
Using $\int u \, dv = uv - \int v \, du$:

$$\int x^2 e^{-2x} dx = x^2 \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right) (2x \, dx)$$ $$= -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} dx$$

Let $u = x \implies du = dx$.
Let $dv = e^{-2x} dx \implies v = -\frac{1}{2}e^{-2x}$.

$$\int x e^{-2x} dx = x \left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right) dx$$ $$= -\frac{1}{2}x e^{-2x} + \frac{1}{2} \int e^{-2x} dx$$ $$= -\frac{1}{2}x e^{-2x} + \frac{1}{2} \left(-\frac{1}{2}e^{-2x}\right) + C$$ $$= -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C$$

$$\int e^{-2x} x^2 dx = -\frac{1}{2}x^2 e^{-2x} + \left( -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} \right) + C$$ $$= -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C$$

Factor out $-\frac{1}{4}e^{-2x}$:

$$\boxed{-\frac{1}{4}e^{-2x} (2x^2 + 2x + 1) + C}$$

Let $u = 4x$, so $du = 4 dx$, which means $dx = \frac{1}{4} du$.

$$\int \cot^5(4x) dx = \frac{1}{4} \int \cot^5(u) du$$

$$\frac{1}{4} \int \cot^5(u) du = \frac{1}{4} \int \cot^3(u) \cot^2(u) du$$ $$= \frac{1}{4} \int \cot^3(u) (\csc^2(u) - 1) du$$ $$= \frac{1}{4} \left( \int \cot^3(u) \csc^2(u) du - \int \cot^3(u) du \right)$$

$\int \cot^3(u) \csc^2(u) du$.
Let $w = \cot(u)$, then $dw = -\csc^2(u) du$.

$$\int w^3 (-dw) = -\frac{w^4}{4} = -\frac{\cot^4(u)}{4}$$

$\int \cot^3(u) du$.

$$\int \cot^3(u) du = \int \cot(u) \cot^2(u) du = \int \cot(u) (\csc^2(u) - 1) du$$ $$= \int \cot(u) \csc^2(u) du - \int \cot(u) du$$

For $\int \cot(u) \csc^2(u) du$, let $w = \cot(u)$, $dw = -\csc^2(u) du$.

$$\int w (-dw) = -\frac{w^2}{2} = -\frac{\cot^2(u)}{2}$$

And $\int \cot(u) du = \ln|\sin(u)|$.
So, $\int \cot^3(u) du = -\frac{\cot^2(u)}{2} - \ln|\sin(u)|$.

$$\frac{1}{4} \left( -\frac{\cot^4(u)}{4} - \left( -\frac{\cot^2(u)}{2} - \ln|\sin(u)| \right) \right) + C$$ $$= \frac{1}{4} \left( -\frac{\cot^4(u)}{4} + \frac{\cot^2(u)}{2} + \ln|\sin(u)| \right) + C$$

Substitute $u = 4x$:

$$\boxed{-\frac{\cot^4(4x)}{16} + \frac{\cot^2(4x)}{8} + \frac{1}{4}\ln|\sin(4x)| + C}$$

Let $u = \sin^{-1}\left(\frac{7x}{3}\right)$ and $dv = dx$.
Then $v = x$.
To find $du$, we differentiate $u$:

$$du = \frac{d}{dx}\left(\sin^{-1}\left(\frac{7x}{3}\right)\right) dx = \frac{\frac{7}{3}}{\sqrt{1 - \left(\frac{7x}{3}\right)^2}} dx$$ $$du = \frac{\frac{7}{3}}{\sqrt{1 - \frac{49x^2}{9}}} dx = \frac{\frac{7}{3}}{\sqrt{\frac{9 - 49x^2}{9}}} dx = \frac{\frac{7}{3}}{\frac{1}{3}\sqrt{9 - 49x^2}} dx = \frac{7}{\sqrt{9 - 49x^2}} dx$$

$$\int \sin^{-1}\left(\frac{7x}{3}\right) dx = x \sin^{-1}\left(\frac{7x}{3}\right) - \int x \frac{7}{\sqrt{9 - 49x^2}} dx$$ $$= x \sin^{-1}\left(\frac{7x}{3}\right) - 7 \int \frac{x}{\sqrt{9 - 49x^2}} dx$$

Let $w = 9 - 49x^2$.
Then $dw = -98x \, dx$, so $x \, dx = -\frac{1}{98} dw$.

$$\int \frac{x}{\sqrt{9 - 49x^2}} dx = \int \frac{1}{\sqrt{w}} \left(-\frac{1}{98}\right) dw = -\frac{1}{98} \int w^{-1/2} dw$$

$$-\frac{1}{49}\sqrt{9 - 49x^2} + C$$

So, the full integral is:

$$x \sin^{-1}\left(\frac{7x}{3}\right) - 7 \left( -\frac{1}{49}\sqrt{9 - 49x^2} \right) + C$$ $$= x \sin^{-1}\left(\frac{7x}{3}\right) + \frac{7}{49}\sqrt{9 - 49x^2} + C$$ $$\boxed{x \sin^{-1}\left(\frac{7x}{3}\right) + \frac{1}{7}\sqrt{9 - 49x^2} + C}$$