1. (a) Given that $2+i$ is a complex root of the cubic polynomial $x^{3}-11 x+20$, determine the other tivo roots (without using a calculator). (b) Hence, (and without using a calculator) determine \[ \int \frac{37}{x^{3}-11 x+20} d x \] (Hint: Use the result of part (a) to write \[ x^{3}-11 x+20=(x-a)\left(x^{2}+b x+c\right) \] for somc $a, b$ and $c$, and usc partial fractions.) (c) (i) Determine the general solution of the differential equation \[ y^{\prime \prime}-\frac{y^{\prime}}{x}=x \sin 11 x \] (Hint: Set $v=y^{\prime}$ and solve the resulting linear differential equation for $v=v(x)$.)

Step 1 ：Since $2+i$ is a root of the cubic polynomial, its complex conjugate $2-i$ is also a root. Thus, we have two roots: $2+i$ and $2-i$.

Step 2 ：Let the third root be $a$. Then, by Vieta's formulas, we have $a+(2+i)+(2-i)=0$, which implies $a=-4$.

Step 3 ：Thus, the cubic polynomial can be factored as $(x-(2+i))(x-(2-i))(x+4) = (x^2-4x+5)(x+4)$.

Step 4 ：Using partial fractions, we can write \(\frac{37}{x^3-11x+20}\) as \(\frac{A}{x+4} + \frac{Bx+C}{x^2-4x+5}\).

Step 5 ：Multiplying both sides by $x^3-11x+20$, we get $37 = A(x^2-4x+5) + (Bx+C)(x+4)$.

Step 6 ：Setting $x=-4$, we get $37 = 21A$, so $A=\frac{37}{21}$.

Step 7 ：Comparing coefficients, we have $B=-\frac{37}{21}$ and $C=\frac{185}{21}$.

Step 8 ：Thus, \(\int \frac{37}{x^3-11x+20} dx = \int \frac{\frac{37}{21}}{x+4} + \frac{-\frac{37}{21}x+\frac{185}{21}}{x^2-4x+5} dx\).

Step 9 ：Integrating, we get \(\frac{37}{21} \ln|x+4| - \frac{37}{21} \int \frac{x}{x^2-4x+5} dx + \frac{185}{21} \int \frac{1}{x^2-4x+5} dx\).

Step 10 ：Completing the square for the denominator of the second integral, we get \(\int \frac{x}{x^2-4x+5} dx = \int \frac{x}{(x-2)^2+1} dx\).

Step 11 ：Using substitution, let $u=x-2$, then $du=dx$. The integral becomes \(\int \frac{u+2}{u^2+1} du\).

Step 12 ：Splitting the fraction, we get \(\int \frac{u}{u^2+1} du + \int \frac{2}{u^2+1} du\).

Step 13 ：Integrating, we get \(\frac{1}{2} \ln(u^2+1) + 2 \arctan(u) + C_1 = \frac{1}{2} \ln((x-2)^2+1) + 2 \arctan(x-2) + C_1\).

Step 14 ：Completing the square for the denominator of the third integral, we get \(\int \frac{1}{x^2-4x+5} dx = \int \frac{1}{(x-2)^2+1} dx\).

Step 15 ：Integrating, we get \(\arctan(x-2) + C_2\).

Step 16 ：Combining all the integrals, we get \(\frac{37}{21} \ln|x+4| - \frac{37}{21} \left(\frac{1}{2} \ln((x-2)^2+1) + 2 \arctan(x-2)\right) + \frac{185}{21} \arctan(x-2) + C\).

Step 17 ：Thus, the final answer is \(\boxed{\frac{37}{21} \ln|x+4| - \frac{37}{42} \ln((x-2)^2+1) + \frac{111}{21} \arctan(x-2) + C}\).