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We need to decompose the integrand into partial fractions. Let $$ \frac{x^2 + 1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3} $$ Multiplying both sides by $(x-1)^2(x+3)$, we get $$ x^2 + 1 = A(x-1)(x+3) + B(x+3) + C(x-1)^2 $$
To find the values of $A$, $B$, and $C$, we can substitute specific values of $x$. Let $x = 1$: $$ 1^2 + 1 = A(0) + B(1+3) + C(0) \implies 2 = 4B \implies B = \frac{1}{2} $$ Let $x = -3$: $$ (-3)^2 + 1 = A(0) + B(0) + C(-3-1)^2 \implies 10 = 16C \implies C = \frac{5}{8} $$ Now, we can substitute any other value of $x$ to find $A$. Let $x = 0$: $$ 0^2 + 1 = A(-1)(3) + B(3) + C(1) \implies 1 = -3A + 3B + C $$ Substituting the values of $B$ and $C$: $$ 1 = -3A + 3\left(\frac{1}{2}\right) + \frac{5}{8} \implies 1 = -3A + \frac{3}{2} + \frac{5}{8} $$ $$ 3A = \frac{3}{2} + \frac{5}{8} - 1 = \frac{12 + 5 - 8}{8} = \frac{9}{8} \implies A = \frac{3}{8} $$ Thus, $A = \frac{3}{8}$, $B = \frac{1}{2}$, and $C = \frac{5}{8}$.
Now we can rewrite the integral as: $$ \int \frac{x^2 + 1}{(x-1)^2(x+3)} dx = \int \left(\frac{3/8}{x-1} + \frac{1/2}{(x-1)^2} + \frac{5/8}{x+3}\right) dx $$ $$ = \frac{3}{8} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{1}{(x-1)^2} dx + \frac{5}{8} \int \frac{1}{x+3} dx $$ $$ = \frac{3}{8} \ln|x-1| + \frac{1}{2} \left(-\frac{1}{x-1}\right) + \frac{5}{8} \ln|x+3| + C $$ $$ = \frac{3}{8} \ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8} \ln|x+3| + C $$
Therefore, the integral is: $$ \int \frac{x^2 + 1}{(x-1)^2(x+3)} dx = \frac{3}{8} \ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8} \ln|x+3| + C $$
Final Answer: $\frac{3}{8} \ln|x-1| - \frac{1}{2(x-1)} + \frac{5}{8} \ln|x+3| + C$
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