Tích phân $\int_{2}^{3}\frac{3x+9}{x^{2}-4x-5}dx$


bài này cũng phải tách nhỉ –  congiola_ktqd 09-11-12 09:49 PM
=3/2. $\int\limits_{2}^{3}$$\frac{2x+6}{x^{2}-4x-5}$.dx= 3/2. $\int\limits_{2}^{3}$$\frac{(2x-4)+10}{x^{2}-4x-5}$.dx= 3/2. $\int\limits_{2}^{3}$$\frac{(x^{2}-4x-5)^{'}}{x^{2}-4x-5}$.dx + 15. $\int\limits_{2}^{3}$$\frac{1}{x^{2}-4x-5}$.dx
 = 3/2. $\ln \left| {x^{2}-4x-5} \right|$ + 15. $\int\limits_{2}^{3}$$\frac{1}{(x-5)(x+1)}$.dx
 =3/2. $\ln \left| {x^{2}-4x-5} \right|$ + 15. $\int\limits_{2}^{3}$$(\frac{1}{(x-5)}-\frac{1}{x+1})$.dx
 =3/2. $\ln \left| {x^{2}-4x-5} \right|$ + 15. ($\ln \left| {x-5} \right|$- $\ln \left| {x+1} \right|$)
 = 3/2 $\ln 8/9$+ 15$\ln 1/2$
Bạn Giang xem lại kết quả nhé, kết quả của bạn Khang là chuẩn rồi đấy –  Đỗ Quang Chính 08-11-12 06:37 PM
Ta có:
      $\int\limits_{2}^{3}\frac{3x+9}{x^{2}-4x-5}dx$
 $=\int\limits_2^3\left(-\frac{4}{5-x}-\frac{1}{x+1}\right)dx$
 $=4\ln(5-x)-\ln(x+1)\left|\begin{array}{l}3\\2\end{array}\right.=-\ln(\frac{27}{4})$
lời giải khó hiểu –  babylionneu 09-11-12 10:12 PM
vote luôn! –  congiola_ktqd 09-11-12 09:49 PM
vote cho bạn –  tanlong838 09-11-12 10:54 AM

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