bai tap tích phân

T

tuyn

a)
[tex]I= \int\limits_{0}^{1}\frac{2x+3}{x^2+ 3x + 5}dx[/tex]
b)
[tex]I= \int\limits_{-1}^{0}\frac{3x + 4}{2x^2 + x - 3}dx[/tex]
a)
[TEX]I= \int_{0}^{1} \frac{d(x^2+3x+5)}{x^2+3x+5}=ln(x^2+3x+5)|_0^1[/TEX]
[TEX]=ln \frac{9}{5}[/TEX]
b)
Ta có: [TEX]2x^2+x-3=(x-1)(2x+3)[/TEX]
Tìm 2 số A,B sao cho:
[TEX]\frac{3x+4}{2x^2+x-3}= \frac{A}{x-1}+ \frac{B}{2x+3}[/TEX]
[TEX]\Rightarrow 3x+4=(2A+B)x+(3A-B) \Rightarrow \left{\begin{2A+B=3}\\{3A-B=4}[/TEX]
[TEX]\Leftrightarrow \left{\begin{A= \frac{7}{5}}\\{B= \frac{1}{5}}[/TEX]
[TEX]I= \int_{-1}^{0}( \frac{7}{5(x-1)}+ \frac{1}{5(2x+3)})dx[/TEX]
[TEX]= ( \frac{5}{5} ln|x-1|+ \frac{1}{10}ln|2x+3|)|_{-1}^{0}[/TEX]
 
Top Bottom