Tính tích phân 0 đến 1 (x+2)dx/(x^2+4x+7)=alnsqrt12+blnsqrt7

Xét \(\int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 4x + 7}}{\rm{d}}x}\)

Đặt \(u = {x^2} + 4x + 7 \Rightarrow du = 2x + 4dx \Rightarrow \frac{1}{2}du = \left( {x + 2} \right)dx\) 

Vậy:

 \(\begin{array}{l} \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 4x + 7}}{\rm{d}}x} = \frac{1}{2}\int\limits_7^{12} {\frac{1}{u}} = \frac{1}{2}\left. {\ln \left| u \right|} \right|_7^{12} = \frac{1}{2}\\ = \frac{1}{2}\ln 12 - \frac{1}{2}\ln 7 = \ln \sqrt {12} - \ln \sqrt 7 . \end{array}\)

Suy ra \(\int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 4x + 7}}{\rm{d}}x} = a\ln \sqrt {12} + b\ln \sqrt 7 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {a = 1}\\ {b = - 1} \end{array}} \right.\)

Vậy a+b=0.

Vậy tổng .