anh puvid ~ chị hangibich giúp em ạ !!!!!! $\to \displaystyle\int\dfrac{x^2}{1}dx$

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1. Đáp án:

   Ta có : 

   $\displaystyle\int\dfrac{x^2}{1}dx$

   $= \displaystyle\int x^2dx$

   $ = \displaystyle\int xxdx$

   `= 1/2x^3 – 1/2`  $\displaystyle\int xxdx$

   `= 1/2x^3 – 1/4 x^3 + 1/4` $\displaystyle\int xxdx$

   `= 1/2x^3 – 1/4x^3 + 1/8x^3 – 1/8` $\displaystyle\int xxdx$

   `…`

   `= \sum_{n=1}^{∞} \frac{(-1)^{n+1}}{2^n}x^3+C`

    `= x^3 \sum_{n=1}^{∞} \frac{(-1)^{n+1}}{2^n}+C`

   `= x^3 ( \sum_{n=0}^{∞} \frac{(-1)^{2n}}{2^{2n+1}}+  \sum_{n=1}^{∞} \frac{(-1)^{2n+1}}{2^{2n}} ) + C`

   `= x^3 (\sum_{n=0}^{∞} \frac{1}{2^{2n+1}}-  \sum_{n=1}^{∞} \frac{1}{2^{2n}}) + C`

   `= x^3 (\frac{1}{2}\sum_{n=0}^{∞} \frac{1}{4^n}-   \frac{1}{4}\sum_{n=0}^{∞} \frac{1}{4^n}) + C`

   `= x^3 (1/4\sum_{n=0}^{∞} \frac{1}{4^n}) + C`

   `= x^3 (1/4 1/(1 – 1/4)) + C`

   `= 1/3 x^3 + C`

    

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2. Đáp án:

   $\dfrac{x^3}{3} + C$

   Giải thích các bước giải:

   $\quad \displaystyle\int \dfrac{x^2}{1}dx$

   $= \displaystyle\int x^2dx$

   $= \dfrac{x^{2+1}}{2+1} + C$

   $=\dfrac{x^3}{3} + C$

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