Quiz 8.5:


Question:

Integrate \( \int \frac{2x^3-4x^2-x-3}{x^2-2x-3}dx \)


Solution:

\( \int \frac{2x^3-4x^2-x-3}{x^2-2x-3}dx = \int 2x dx + \int \frac{A}{x+1}+\frac{B}{x-3}dx \)


\( \frac{5x-3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3} \)
\( 5x-3 = A(x-3)+B(x+1) \)
Letting \( x=3 \), we get that \( 4B=12 \) or \( B=3 \)
Letting \( x=-1 \), we get that \(-4A=-8 \) or \( A=2 \)

\( = \int 2x dx + \int \left ( \frac{2}{x+1}+\frac{3}{x-3} \right ) dx \)
\( = x^2 +2 \ln |x+1| +3\ln |x-3| + C \)
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