tich phân

Question

$\int\limits_{2}^{3}\frac{xdx}{(x-1)^{2}(x+1)^{3}}$

score 0 · Answer 1

Ta phân tích như sau:

$\frac{x}{{{{(x - 1)}^2}{{(x + 1)}^3}}} = \frac{A}{{{{(x - 1)}^2}}} + \frac{B}{{{{(x + 1)}^3}}} + \frac{C}{{(x - 1)(x + 1)}}$

$= \frac{{A({x^3} + 3{x^2} + 3x + 1) + B({x^2} - 2x + 1) + C({x^3} + {x^2} - x - 1)}}{{{{(x - 1)}^2}{{(x + 1)}^3}}}$

Từ đây, ta được hệ phương trình:

$\begin{cases}A+C=0 \\ 3A+B+C=0 \\ 3A-2B-C=1 \\ A+B-C=0 \end{cases}$

$\Leftrightarrow A = \frac{1}{8},B = - \frac{1}{4},C = - \frac{1}{8}$

Khi đó, ta có:

$I = \frac{1}{8}\int\limits_2^3 {\frac{{dx}}{{{{(x - 1)}^2}}}} - \frac{1}{4}\int\limits_2^3 {\frac{{dx}}{{{{(x + 1)}^3}}}} - \frac{1}{8}\int\limits_2^3 {\frac{{dx}}{{(x - 1)(x + 1)}}}$

$= \left[ { - \frac{1}{8}\frac{1}{{x - 1}} + \frac{1}{8}\frac{1}{{{{(x + 1)}^2}}}} \right]\begin{cases}3 \\ 2 \end{cases} - \frac{1}{{16}}\int\limits_2^3 {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right)dx}$