Answer:
First, express the fraction in partial fractions
Write it out as an identity.
As the denominator has a repeated linear factor, the power of the repeated factor tells us the number of times the factor should appear in the partial fraction. A factor that is squared in the original denominator will appear in the denominator of two of the partial fractions - once squared and once just as it is:
[tex]\dfrac{2x^2+x+1}{(x+3)(x-1)^2} \equiv \dfrac{A}{(x+3)}+\dfrac{B}{(x-1)}+\dfrac{C}{(x-1)^2}[/tex]
Add the partial fractions:
[tex]\dfrac{2x^2+x+1}{(x+3)(x-1)^2} \equiv \dfrac{A(x-1)^2+B(x+3)(x-1)+C(x+3)}{(x+3)(x-1)^2}[/tex]
Cancel the denominators from both sides of the original identity, so the numerators are equal:
[tex]2x^2+x+1 \equiv A(x-1)^2+B(x+3)(x-1)+C(x+3)[/tex]
Now solve for A and C by substitution.
Substitute values of x which make one of the expressions equal zero (to eliminate all but one of A, B and C):
[tex]\begin{aligned}x=1 \implies 2(1)^2+(1)+1 & = A(1-1)^2+B(1+3)(1-1)+C(1+3)\\4 & = 4C\\\implies C & =1\end{aligned}[/tex]
[tex]\begin{aligned}x=-3 \implies 2(-3)^2+(-3)+1 & = A(-3-1)^2+B(-3+3)(-3-1)+C(-3+3)\\16 & = 16A\\\implies A & =1\end{aligned}[/tex]
Find B by comparing coefficients:
[tex]\begin{aligned}2x^2+x+1 & = A(x^2-2x+1)+B(x^2+2x-3)+C(x+3)\\ & = (A+B)x^2+(2B-2A+C)x+(A-3B+3C)\end{aligned}[/tex]
[tex]\begin{aligned}\implies 2x^2 & = (A+B)x^2\\2 & = A + B\\\textsf{substituting found value of A}: \quad2 & = 1+B\\\implies B & = 1 \end{aligned}[/tex]
Replace the found values of A, B and C in the original identity:
[tex]\implies \dfrac{2x^2+x+1}{(x+3)(x-1)^2} \equiv \dfrac{1}{(x+3)}+\dfrac{1}{(x-1)}+\dfrac{1}{(x-1)^2}[/tex]
Now integrate:
[tex]\begin{aligned}\displaystyle \int \dfrac{2x^2+x+1}{(x+3)(x-1)^2}\:dx &=\int \dfrac{1}{(x+3)}+\dfrac{1}{(x-1)}+\dfrac{1}{(x-1)^2}\:\:dx\\\\& =\int \dfrac{1}{(x+3)}\:dx\:\:+ \int\dfrac{1}{(x-1)}\:dx\:\:+\int \dfrac{1}{(x-1)^2}\:dx\\\\& = \ln |x+3|+ \ln |x-1|-\dfrac{1}{x-1}+C\end{aligned}[/tex]
