BÀI TẬP TÍCH PHÂN BA THAM SỐ a, b, c có hướng dẫn Casio

Biết $\int\limits_{0}^{3}{\frac{\text{d}x}{(x+2)(x+4)}\text{d}x}=a\ln 2+b\ln 5+c\ln 7\,(a,b,c\in \mathbb{Q}).$ Giá trị của biểu thức \[2a+3b-c\] bằng

A. $5.$                                        B. $4.$                                     C. $2.$                                     D. $3.$

Lời giải

Chọn D

Cách 1: Ta $\underset{0}{\overset{3}{\mathop \int }}\,\frac{dx}{\left( x+2 \right)\left( x+4 \right)}=\frac{1}{2}\underset{0}{\overset{3}{\mathop \int }}\,\frac{\left[ \left( x+4 \right)-\left( x+2 \right) \right]dx}{\left( x+2 \right)\left( x+4 \right)}=\frac{1}{2}\underset{0}{\overset{3}{\mathop \int }}\,\left( \frac{1}{x+2}-\frac{1}{x+4} \right)dx$

 $=\frac{1}{2}\text{ln }\!\!~\!\!\text{ }\left| \frac{x+2}{x+4} \right|_{0}^{3}=\frac{1}{2}\text{ln }\!\!~\!\!\text{ }\frac{5}{7}-\frac{1}{2}\text{ln }\!\!~\!\!\text{ }\frac{1}{2}=\frac{1}{2}\text{ln }\!\!~\!\!\text{ }2+\frac{1}{2}\ln 5-\frac{1}{2}\ln 7=a\text{ }\!\!~\!\!\text{ ln }\!\!~\!\!\text{ }2+b\text{ }\!\!~\!\!\text{ ln }\!\!~\!\!\text{ }5+c\text{ }\!\!~\!\!\text{ ln }\!\!~\!\!\text{ }7$.

Do $a,b,c\in \mathbb{Q}$ nên $a=\frac{1}{2};b=\frac{1}{2};c=-\frac{1}{2}$. Vây $2a+3b-c=1+\frac{3}{2}+\frac{1}{2}=3.$

Cách 2: Sử dụng MTCT

Từ giả thiết, ta có:

$\underset{0}{\overset{3}{\mathop \int }}\,\frac{dx}{\left( x+2 \right)\left( x+4 \right)}=a\text{ln }\!\!~\!\!\text{ }2+b\text{ln }\!\!~\!\!\text{ }5+c\ln 7=\text{ln }\!\!~\!\!\text{ }\left( {{2}^{a}}{{.5}^{b}}{{.7}^{c}} \right)$ $\Rightarrow {{2}^{a}}{{.5}^{b}}{{.7}^{c}}={{e}^{\underset{{{e}^{0}}}{\overset{3}{\mathop \int }}\,\frac{dx}{\left( x+2 \right)\left( x+4 \right)}}}=\sqrt{\frac{10}{7}}=\frac{\sqrt{2}.\sqrt{5}}{\sqrt{7}}={{2}^{\frac{1}{2}}}{{.5}^{\frac{1}{2}}}{{.7}^{-\frac{1}{2}}}$.

Do $a,b,c\in \mathbb{Q}$ nên $a=\frac{1}{2};b=\frac{1}{2};c=-\frac{1}{2}$. Vây $2a+3b-c=1+\frac{3}{2}+\frac{1}{2}=3.$