tính nguyên hàm I_0= $\int{\dfrac{1}{(x-1)(x+3)^3}}dx$

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1. Đáp án:

   $I_o=\dfrac{x+5}{16(x+3)^2}+ \dfrac{1}{64}\ln\left|\dfrac{x-1}{x+3}\right| + C$

   Giải thích các bước giải:

   $\begin{array}{l}\quad I_o=\displaystyle\int\dfrac{1}{(x-1)(x+3)^3}dx\\ = \displaystyle\int\left[-\dfrac{1}{4(x+3)^3} – \dfrac{1}{16(x+3)^2} – \dfrac{1}{64(x+3)} + \dfrac{1}{64(x-1)}\right]dx\\ = -\dfrac{1}{4}\displaystyle\int\dfrac{dx}{(x+3)^3}-\dfrac{1}{16}\displaystyle\int\dfrac{dx}{(x+3)^2} – \dfrac{1}{64}\displaystyle\int\dfrac{dx}{x+3} + \dfrac{1}{64}\displaystyle\int\dfrac{dx}{x-1}\\ = -\dfrac{1}{4}\displaystyle\int\dfrac{d(x+3)}{(x+3)^3}-\dfrac{1}{16}\displaystyle\int\dfrac{d(x+3)}{(x+3)^2} – \dfrac{1}{64}\displaystyle\int\dfrac{d(x+3)}{x+3} + \dfrac{1}{64}\displaystyle\int\dfrac{d(x-1)}{x-1}\\ = -\dfrac14\cdot\left[-\dfrac{1}{2(x+3)^2}\right] -\dfrac{1}{16}\cdot\left(-\dfrac{1}{x+3}\right) – \dfrac{1}{64}\ln|x+3| + \dfrac{1}{64}\ln|x-1| + C\\ = \dfrac{1}{8(x+3)^2} +\dfrac{1}{16(x+3)} + \dfrac{1}{64}\ln\left|\dfrac{x-1}{x+3}\right| + C\\ = \dfrac{x+5}{16(x+3)^2}+ \dfrac{1}{64}\ln\left|\dfrac{x-1}{x+3}\right| + C\\ \end{array}$

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2. Đáp án:

   $I=\dfrac{1}{64}ln\left | \dfrac{x-1}{x+3} \right |+\dfrac{x+5}{16(x+3)^2}+C$

   Giải thích các bước giải:

    $I=\int \left(\dfrac{1}{64}·\dfrac{1}{x-1}-\dfrac{1}{64}·\dfrac{1}{x+3}-\dfrac{1}{16}·\dfrac{1}{(x+3)^2}-\dfrac{1}{4}·\dfrac{1}{(x+3)^3} \right)dx$

   $⇒I=\dfrac{1}{64}ln\left | \dfrac{x-1}{x+3} \right |+\dfrac{1}{16}·\dfrac{1}{x+3}+\dfrac{1}{8}·\dfrac{1}{(x+3)^2}+C$

   $⇒I=\dfrac{1}{64}ln\left | \dfrac{x-1}{x+3} \right |+\dfrac{x+5}{16(x+3)^2}+C$

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